# NCERT Solutions For Class 12 Maths Chapter 7

## NCERT Solutions Class 12 Maths Integrals Rational Numbers

### Ncert Solutions For Class 12 Maths Chapter 7 PDF Free Download

NCERT solutions for class 12 maths chapter 7 can be accessed here. The NCERT maths book of class 12 includes the concept of integrals in chapter 7. In this chapter students learn about integral calculus (definite and indefinite), its formulas and their properties. This topics is extremely important for both CBSE board exam and for competitive exams like JEE and NEET.

NCERT class 12 maths solutions for chapter 7 integrals given here are detailed and easy to understand. These simple class 12 maths NCERT solutions for integrals are very simple and can help the students to understand them very easily. The class 12 maths chapter 7 PDF is also available for download.

The NCERT solutions is one of the best tool to prepare physics for class 12 Class 12 Maths Integrals Rational Numbers is a crucial chapter in CBSE class 12. Students must prepare this chapter well to score well in their examination. The NCERT Solutions which contain solved math exercises is given below so that students can understand the concepts of this chapter in depth.

### NCERT Solutions Class 12 Maths Chapter 7 Exercises

Exercise – 7.1

Question 1:

By the method of inspection obtain an integral (or anti – derivative) of the sin 3x.

As the derivative is sin 3x and x is the function of the anti – derivative of sin 3x.

$\frac{d}{dx} (cos\; 3x) = – 3 sin\; 3x \\ sin\; 3x = – \frac{1}{3} \frac{d}{dx} (cos\; 3x) \\ sin\; 3x = \frac{d}{dx} (- \frac{1}{3} cos\; 3x) \\ Hence,\; the\; anti – derivative\; of\; sin\; 3x \;is\; (- \frac{1}{3} cos\; 3x)$

Question 2:

By the method of inspection obtain an integral (or anti – derivative) of the cos 2x.

As the derivative is cos 2x and x is the function of the anti – derivative of cos 2x

$\frac{d}{dx} (sin\; 2x) = – 2 cos\; 2x \\ cos\; 2x = \frac{1}{2} \frac{d}{dx} (sin\; 2x) \\ cos\; 2x = \frac{d}{dx} (\frac{1}{2} (sin\; 2x)) \\ Hence,\; the\; anti – derivative\; of\; sin\; 2x \;is\; (- \frac{1}{2} cos\; 2x)$

Question 3:

By the method of inspection obtain an integral (or anti – derivative) of the e5x.

As the derivative is e5x and x is the function of the anti – derivative of e5x

$\frac{d}{dx} (e ^{5x}) = 5 e ^{5x} \\ e ^{5x} = \frac{1}{5} \frac{d}{dx} (e^{5x}) \\ e ^{5x} = \frac{d}{dx} (\frac{1}{5} e^{5x}) \\ Hence,\; the\; anti – derivative\; of\; e ^{5x} is \frac{1}{5} e^{5x}$

Question 4:

By the method of inspection obtain an integral (or anti – derivative) of the (mx  + n) 2.

As the derivative is (mx  + n) 2 and x is the function of the anti – derivative of (mx  + n) 2

$\frac{d}{dx} (mx + n)^{3} = 3m (mx + n) ^{2}\\ (mx + n) ^{2} = \frac{1}{3m} \frac{d}{dx} (mx + n) ^{3} \\ (mx + n) ^{2} = \frac{d}{dx} (\frac{1}{3m} (mx + n) ^{3}) \\ Hence,\; the\; anti – derivative\; of\; (mx + n) ^{2} is \frac{1}{3m} (mx + n) ^{3}$

Question 5:

By the method of inspection obtain an integral (or anti – derivative) of the sin 3x – 5 e 2x

As the derivative is (sin 3x – 5 e 2x) and x is the function of the anti – derivative of (sin 3x – 5 e 2x)

$\frac{d}{dx} (- \frac{1}{3} cos\; 3x – \frac{5}{2} e^{2x}) = sin 3x – 5 e^{2x} \\ Hence,\; the\; anti – derivative\; of\; sin\; 3x – 5 e^{2x} \;is\; (- \frac{1}{3} cos\; 3x – \frac{5}{2} e^{2x})$

Question 6:

By the method of inspection obtain an integral of the $\int (4 e^{2u} + 1) du$

Integral of $(4 e^{2u} + 1)$ and u is the function of the integral $(4 e^{2u} + 1)$.

$\int (4 e^{2u} + 1) du \\ 4 \int e^{2u} du + \int 1 du \\ 4 (\frac{e^{2u}}{2}) + u + c \\ 2 e^{2u} + u + c \\ Where\; c\; is\; the\; constant.$

Question 7:

By the method of inspection obtain an integral of the $\int u^{2} (1 – \frac{1}{u^{2}}) du$

Integral of $u^{2} (1 – \frac{1}{u^{2}})$ and u is the function of the integral $u^{2} (1 – \frac{1}{u^{2}})$ $\int u^{2} (1 – \frac{1}{u^{2}}) du \\ \int (u^{2} – 1) du \\ \frac{u^{3}}{3} – u + c \\ Where\; c\; is\; the\; constant$

Question 8:

By the method of inspection obtain an integral of the $\int (a u^{2} + b u + c) du$

Integral of $a u^{2} + b u + c$ and u is the function of the integral $a u^{2} + b u + c$ $\int (a u^{2} + b u + c) du \\ a \int (u^{2}) du + b \int u du + c \int 1 du \\ a (\frac{u^{3}}{3}) + b (\frac{u^{2}}{2}) + cu + C \\ \\ Where\; C\; is\; the\; constant$

Question 9:

By the method of inspection obtain an integral of the $\int (a u^{2} + e^{u}) du$

Integral of $a u^{2} + e^{u}$ and u is the function of the integral $a u^{2} + e^{u}$ $\int (a u^{2} + e^{u}) du \\ a \int (u^{2}) du + \int e^{u} du \\ a (\frac{u^{3}}{3}) + e^{u} + C \\ \\ Where\; C\; is\; the\; constant$

Question 10:

By the method of inspection obtain an integral of the $\int (\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2} du$

Integral of $(\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2}$ and u is the function of the integral $(\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2}$ $(\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2} \\ \int (u + \frac{1}{u} – 2) du \\ \int u du + \int \frac{1}{u} du – 2 \int 1 du \\ \frac{u^{2}}{2} + log \left | u \right | – 2 u + C \\ Where\; C\; is\; the\; constant$

Question 11:

By the method of inspection obtain an integral of the $\int \frac{u^{3} + 4 u^{2} + 4}{u ^{2}} du$

Integral of and u is the function of the integral $\frac{u^{3} + 4 u^{2} + 4}{u ^{2}}$ $\int \frac{u^{3} + 4 u^{2} + 4}{u ^{2}} du \\ \int u du + 4 \int 1 du + \int \frac{4}{u^{2}} du \\ \frac{u ^{2}}{2} + 4 u + \frac{4}{x} + C \\ Where\; C\; is\; the\; constant$

Question 12:

By the method of inspection obtain an integral of the $\frac{u^{3} + 4 u + 4}{\sqrt{u}}$

Integral of $\frac{u^{3} + 4 u + 4}{\sqrt{u}}$ and u is the function of the integral $\frac{u^{3} + 4 u + 4}{\sqrt{u}}$ $\int \frac{u^{3} + 4 u + 4}{\sqrt{u}} du \\ \int (u ^{\frac{5}{2}} + 4 u ^{\frac{1}{2}} + 4 u^{- \frac{1}{2}}) \\ = \frac{u ^{\frac{7}{2}}}{\frac{7}{2}} + \frac{4 (u ^{\frac{3}{2}})}{\frac{3}{2}} + \frac{4 (u ^{\frac{1}{2}})}{\frac{1}{2}} + C \\ = \frac{2}{7} (u ^{\frac{7}{2}}) + \frac{8}{3} (u ^{\frac{3}{2}}) + 8 u ^{\frac{1}{2}} + C \\ = \frac{2}{7} (u ^{\frac{7}{2}}) + \frac{8}{3} (u ^{\frac{3}{2}}) + 8 \sqrt{u} + C \\ Where\; C\; is\; the\; constant$

Question 13:

By the method of inspection obtain an integral of the $\frac{u^{3} – u^{2} + u + 1}{u – 1}$

Integral of $\frac{u^{3} – u^{2} + u + 1}{u – 1}$ and u is the function of the integral $\frac{u^{3} – u^{2} + u + 1}{u – 1}$ $\int \frac{u^{3} – u^{2} + u + 1}{u – 1} du \\ On\; divinding,\; we\; get\; \\ \int (u^{2} + 1) du \\ \int u^{2} du + \int 1 du \\ \frac{u^{3}}{3} + u + C Where\; C\; is\; the\; constant$

Question 14:

By the method of inspection obtain an integral of the $(1 – u) \sqrt{u}$

Integral of $(1 – u) \sqrt{u}$ and u is the function of the integral $(1 – u) \sqrt{u}$ $\int (1 + u) \sqrt{u}\; du \\ \int (\sqrt{u} + u^{\frac{3}{2}}) du \\ \int u^{\frac{1}{2}} du + \int u^{\frac{3}{2}} du \\ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + \frac{u^{\frac{5}{2}}}{\frac{5}{2}} + C \\ \frac{2}{3} u^{\frac{3}{2}} + \frac{2}{5} u^{\frac{5}{2}} + C \\ Where\; C\; is\; the\; constant$

Question 15:

By the method of inspection obtain an integral of the $\sqrt{u} (3u^{2} + 2u + 5)$

Integral of $\sqrt{u} (3u^{2} + 2u + 5)$ and u is the function of the integral $\sqrt{u} (3u^{2} + 2u + 5)$ $\int \sqrt{u} (3u^{2} + 2u + 5) du \\ \int (3u ^{\frac{5}{2}} + 2u ^{\frac{3}{2}} + 5u ^{\frac{1}{2}}) du \\ 3 \int u ^{\frac{5}{2}} du + 2 \int u ^{\frac{3}{2}} du + 5 \int u ^{\frac{1}{2}} du \\ 3 (\frac{u ^{\frac{7}{2}}}{\frac{7}{2}}) + 2 (\frac{u ^{\frac{5}{2}}}{\frac{5}{2}}) + 5 (\frac{u ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ \frac{6}{7} u ^{\frac{7}{2}} + \frac{4}{5} u ^{\frac{5}{2}} + \frac{10}{3} u ^{\frac{3}{2}} + C \\ Where\; C\; is\; the\; constant$

Question 16:

By the method of inspection obtain an integral of the $2 u – 2 cos\; u + e ^{u}$

Integral of $2 u – 2 cos\; u + e ^{u}$ and u is the function of the integral $2 u – 2 cos\; u + e ^{u}$ $\int (2 u – 2 cos\; u + e ^{u}) du \\ 2 \int u du – 2 \int cos\; u du + \int e ^{u} du \\ 2 \frac{u ^{2}}{2} – 2 (sin u) + e ^{u} + C \\ u ^{2} – 2 sin\; u + e ^{u} + C Where\; C\; is\; the\; constant$

Question 17:

By the method of inspection obtain an integral of the $(4 v^{2} + 2 sin v + 6 \sqrt{v})$

Integral of $(4 v^{2} + 2 sin v + 6 \sqrt{v})$ and v is the function of the integral $(4 v^{2} + 2 sin v + 6 \sqrt{v})$ $\int (4 v^{2} + 2 sin v + 6 \sqrt{v})\; dv \\ 4 \int v^{2} dv + 2 \int sin v dv + 6 \int v^{\frac{1}{2}} \\ \frac{4 v^{3}}{3} + 2 (- cos\; v) + 6 (\frac{v ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ \frac{4}{3} v^{3} – 2 cos\; v + 4 v ^{\frac{3}{2}} + C \\ Where\; C\; is\; the\; constant$

Question 18:

By the method of inspection obtain an integral of the $sec\; \Theta (tan\; \Theta + sec\; \Theta)$

Integral of $sec\; \Theta (tan\; \Theta + sec\; \Theta)$ and $\Theta$ is the function of the integral $sec\; \Theta (tan\; \Theta + sec\; \Theta)$ $\int sec\; \Theta (tan\; \Theta + sec\; \Theta) d\Theta \\ \int (sec\; \Theta\; tan\; \Theta + sec ^{2}\; \Theta) d\Theta \\ sec\; \Theta\; + tan\; \Theta + K \\ Where\; K\; is\; the\; constant$

Question 19:

By the method of inspection obtain an integral of the $\frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta}$

Integral of $\frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta}$ and $\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}$c is the function of the integral $\frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta}$ $\int \frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta} d\Theta \\ \int \frac{\frac{1}{cos ^{2}\; \Theta}}{\frac{1}{sin ^{2}\; \Theta}} d\Theta \\ \int \frac{sin ^{2}\; \Theta}{cos ^{2}\; \Theta} d\Theta \\ \int (tan ^{2}\; \Theta) d\Theta \\ \int (sec ^{2}\; \Theta – 1) d\Theta \\ \int sec ^{2}\; \Theta d\Theta – \int 1 d\Theta \\ tan \Theta – \Theta + K \\ Where\; K\; is\; the\; constant$

Question 20:

By the method of inspection obtain an integral of the $\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}$

Integral of $\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}$ and $\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}$ is the function of the integral $\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}$ $\int \frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta} d\Theta \\ \int (\frac{3}{cos ^{2}\; \Theta} – \frac{2 sin\; \Theta}{cos ^{2}\; \Theta}) d\Theta \\ 3 \int sec ^{2}\; \Theta\; d\Theta – 2 \int tan\; \Theta \; sec\; \Theta d\Theta \\ 3 tan\; \Theta – 2 sec\; \Theta + K \\ Where\; K\; is\; the\; constant$

Question 21:

Which of the following below is an integral of $\sqrt{u} + \frac{1}{\sqrt{u}}$:

$(a) \frac{1}{3} u^{\frac{1}{3}} + 2 u^{\frac{1}{2}} + C \\ (b) \frac{2}{3} u^{\frac{2}{3}} + \frac{1}{2} u^{2} + C \\ (c) \frac{2}{3} u^{\frac{3}{2}} + 2 u^{\frac{1}{2}} + C \\ (d) \frac{3}{2} u^{\frac{3}{2}} + \frac{1}{2} u^{\frac{1}{2}} + C$

Integral of $\sqrt{u} + \frac{1}{\sqrt{u}}$ and u is the function of the integral $\sqrt{u} + \frac{1}{\sqrt{u}}$ $\int \sqrt{u} + \frac{1}{\sqrt{u}} du \\ \int u ^{\frac{1}{2}} du + \int u^{- \frac{1}{2}} du \\ \frac{u ^{\frac{3}{2}}}{\frac{3}{2}} + \frac{u ^{\frac{1}{2}}}{\frac{1}{2}} + C \\ \frac{3}{2} u ^{\frac{3}{2}} + 2 u ^{\frac{1}{2}} \\ Option\; c\; is\; correct$

Question 22:

$Suppose\; \frac{d}{dr} f (r) = 4 r^{3} – \frac{3}{r^{4}},\; in\; such\; a\; way\; that\; f (2) = 0,\; then\; f (r)\; is\\ (a) r^{4} + \frac{1}{r ^{3}} – \frac{129}{8} (b) r^{3} + \frac{1}{r ^{4}} + \frac{129}{8} (c) r^{4} + \frac{1}{r ^{3}} + \frac{129}{8} (d) r^{3} + \frac{1}{r ^{4}} – \frac{129}{8}$

Given,

$\frac{d}{dr} f (r) = 4 r^{3} – \frac{3}{r^{4}} \\ Integral\; of\; 4 r^{3} – \frac{3}{r^{4}} = f (r) \\ f (r) = \int 4 r^{3} – \frac{3}{r^{4}}\; dr \\ f (r) = 4 \int r^{3} dr – 3 \int (r ^{- 4}) dr \\ f (r) = 4 \frac{r ^{4}}{4} – 3 \frac{r ^{- 3}}{- 3} + K \\ f (r) = r ^{4} + \frac{1}{r ^{3}} + K \\ And, \\ f (2) = 0 \\ f (2) = 2 ^{4} + \frac{1}{2 ^{3}} + K = 0 \\$ $16 + \frac{1}{8} + K = 0 \\ K = – \frac{129}{8} \\ f (r) = r ^{4} + \frac{1}{r ^{3}} – \frac{129}{8} \\ Option\; (a)\; is\; correct$

Exercise 7.2

Question 1:

Obtain an integral (or anti – derivative) of the $\frac{2 u}{1 + u^{2}}$

Suppose, 1 + u 2 = z

2u du = dz

$\int \frac{2u}{1 + u^{2}} = \int \frac{1}{z}\; dz \\ log \left | z \right | + K \\ log \left | 1 + u^{2} \right | + K \\ log (1 + u^{2}) + K$

Question 2:

Obtain an integral (or anti – derivative) of the $\frac{(log\; u) ^{2}}{u}$

Suppose, $log \left | u \right | = z$ $log \left | u \right | = z \\ \frac{1}{u} du = dz \\ \int \frac{(log \left | u \right |)^{2}}{u} du = \int z^{2} dz \\ = \frac{z ^{3}}{3} + C \\ = \frac{(log \left | u \right |) ^{3}}{3} + C$

Question 3:

Obtain an integral (or anti – derivative) of the $\frac{1}{u + u\; log\; u}$

$\frac{1}{u + u\; log\; u} = \frac{1}{u (1 + log\; u)}$

Suppose, 1 + log u = z

$\frac{1}{u} du = dz \\ \int \frac{1}{u (1 + log\; u)} du = \int \frac{1}{z} dz = log \left | z \right | + C \\ = log \left | 1 + log u \right | + C$

Question 4:

Obtain an integral (or anti – derivative) of the $sin\; u . sin (cos\; u)$

$sin\; u . sin (cos\; u)$

Suppose, cos u = x

– sin u du = dx

$\int sin\; u . sin (cos\; u) du = – \int sin\; x dx \\ = – [- cos x] + C \\ = cos x + C \\ = cos (cos u) + C$

Question 5:

Obtain an integral (or anti – derivative) of the $sin\; (mr + n) cos\; (mr + n)$

Suppose, $sin\; (mr + n) cos\; (mr + n) = \frac{2 sin\; (mr + n) cos\; (mr + n)}{2} = \frac{sin 2 (mr + n)}{2} \\ Suppose\; 2 (mr + n) = z \\ 2 m dr = dz \\ \int \frac{sin 2 (mr + n)}{2} dr = \frac{1}{2} \int \frac{sin\; z\; dz}{2m} \\ = \frac{1}{4m} [- cos z] + C \\ = – \frac{1}{4m} cos 2 (mr + n) + C$

Question 6:

Obtain an integral (or anti – derivative) of the $\sqrt{mr + n}$

Suppose, mr + n = z

m dr = dz

$dr = \frac{1}{m} dz \\ \int (mr + n)^{\frac{1}{2}} dr = \frac{1}{m} \int z ^{\frac{1}{2}} dz \\ \frac{1}{m} (\frac{z ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ \frac{2}{3m} (mr + n)^{\frac{3}{2}} + C$

Question 7:

Obtain an integral (or anti – derivative) of the $u \sqrt{u + 2}$

Suppose, u + 2 = z

du = dz

$\int u \sqrt{u + 2} du = \int (z – 2) \sqrt{z} dz \\ = \int (z ^{\frac{3}{2}} – 2z ^{\frac{1}{2}}) dz \\ = \int z ^{\frac{3}{2}} dz – 2 \int z ^{\frac{1}{2}}) dz \\ = \frac{z ^{\frac{5}{2}}}{\frac{5}{2}} – 2 \frac{z ^{\frac{3}{2}}}{\frac{3}{2}} + C \\ = \frac{2}{5} z ^{\frac{5}{2}} – \frac{4}{3} z ^{\frac{3}{2}} + C \\ = \frac{2}{5} (x + 2) ^{\frac{5}{2}} – \frac{4}{3} (x + 2) ^{\frac{3}{2}} + C \\$

Question 8:

Obtain an integral (or anti – derivative) of the $u \sqrt{1 + 2 u ^{2}}$

Suppose, 1 + 2 u2 = z

4u du = dz

$\int u \sqrt{1 + 2 u ^{2}} du = \int \frac{\sqrt{z}}{4} dz \\ = \frac{1}{4} \int z ^{\frac{1}{2}} dz \\ = \frac{1}{4} (\frac{z ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ = \frac{1}{6} (1 + 2 u^{2}) ^{\frac{3}{2}} + C$

Question 9:

Obtain an integral (or anti – derivative) of the $(4 u + 2) \sqrt{u ^{2} + u + 1}$

Suppose, u 2 + u + 1 = z

(2u + 1) du = dz

$\int (4 u + 2) \sqrt{u ^{2} + u + 1} du = \int 2 \sqrt{z} dz \\ = 2 \int \sqrt{z} dz \\ = 2 (\frac{z ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ = \frac{4}{3} (u ^{2} + u + 1) ^{\frac{3}{2}} + C$

Question 10:

Obtain an integral (or anti – derivative) of the $\frac{1}{u – \sqrt{u}}$

$\frac{1}{u – \sqrt{u}} = \frac{1}{\sqrt{u} (\sqrt{u} – 1)}$

Suppose,

$\sqrt{u} – 1 = z \frac{1}{2 \sqrt{u}} du = dz \\ \int \frac{1}{\sqrt{u} (\sqrt{u} – 1)} du = \int \frac{2}{z} dz \\ 2 log \left | z \right | + C \\ 2 log \left | \sqrt{u} – 1 \right | + C$

Question 11:

Obtain an integral (or anti – derivative) of the $\frac{u}{\sqrt{u + 4}},$ x > 0

Suppose, u + 4 = r

du = dr

$\int \frac{u}{\sqrt{u + 4}} du = \int \frac{(r – 4)}{\sqrt{r}} dr \\ = \int (\sqrt{r} – \frac{4}{\sqrt{r}}) dr \\ = \frac{r ^{\frac{3}{2}}}{\frac{3}{2}} – 4 (\frac{r ^{\frac{1}{2}}}{\frac{1}{2}}) + C \\ = \frac{2}{3} r ^{\frac{3}{2}} – 8 r ^{\frac{1}{2}} + C \\ = \frac{2}{3} r. r ^{\frac{1}{2}} – 8 r ^{\frac{1}{2}} + C \\ = \frac{2}{3} r ^{\frac{1}{2}} (r – 12) + C \\ = \frac{2}{3} (u + 4) ^{\frac{1}{2}} (u + 4 – 12) + C \\ = \frac{2}{3} \sqrt{(u + 4)} (u – 8) + C$

Question 12:

Obtain an integral (or anti – derivative) of the $(u ^{3} – 1) ^{\frac{1}{3}} u ^{5}$

Suppose, u 3 – 1 = r

3 u 2 = dr

$\int (u ^{3} – 1) ^{\frac{1}{3}} u ^{5} du = \int (u ^{3} – 1) ^{\frac{1}{3}} u ^{3} . u ^{2} du \\ = \int r ^{\frac{1}{3}} (r + 1) \frac{dr}{3} \\ = \frac{1}{3} \int (r ^{\frac{4}{3}} + r ^{\frac{1}{3}}) dr \\ = \frac{1}{3} [\frac{r ^{\frac{7}{3}}}{\frac{7}{3}} + \frac{r ^{\frac{4}{3}}}{\frac{4}{3}}] + C \\ = \frac{1}{3} [\frac{3}{7} r ^{\frac{7}{3}} + \frac{3}{4} r ^{\frac{4}{3}}] + C \\ = \frac{1}{7} (u ^{3} – 1) ^{\frac{7}{3}} + \frac{1}{4} (u ^{3} – 1) ^{\frac{4}{3}}] + C$

Question 13:

Obtain an integral (or anti – derivative) of the $\frac{u ^{2}}{(2 + 3u ^{3}) ^{3}} \\$

Suppose, $2 + 3u ^{3} = z \\ 9 u ^{2} du = dz \\ \int \frac{u ^{2}}{(2 + 3 u ^{3})} du = \frac{1}{9} \int \frac{dz}{(z) ^{3}} \\ = \frac{1}{9} \int {(z) ^{- 3}} dz \\ = \frac{1}{9} (\frac{z ^{- 2}}{- 2}) + C \\ = – \frac{1}{18} (\frac{1}{z ^{2}}) + C \\ = \frac{- 1}{18 (2 + 3u ^{3}) ^{2}} + C \\$

Question 14:

Obtain an integral (or anti – derivative) of the $\frac{1}{u (log u) ^{n}}, x > 0 \\$

Suppose,  $log u = z \\ \frac{1}{u} du = dz \\ \int \frac{1}{u (log u) ^{n}} du = \int \frac{dz}{z ^{n}} \\ = \int z ^{- n} dz \\ = \frac{z ^{- n + 1}}{- n + 1} + C \\ = \frac{z ^{1 – n}}{1 -n} + C \\ = \frac{x ^{1 – n}}{1 -n} + C$

Question 15:

Obtain an integral (or anti – derivative) of the $\frac{u}{9 – 4 u ^{2}}$

Suppose, $9 – 4 u ^{2} = r \\ – 8 u du = dr \\ \int \frac{u}{9 – 4 u ^{2}} = – \frac{1}{8} \int \frac{1}{r} dr \\ = – \frac{1}{8} log \left | r \right | + C \\ = – \frac{1}{8} log \left | 9 – 4 u ^{2} \right | + C$

Question 16:

Obtain an integral (or anti – derivative) of the $e ^{2 m + 3}$

Suppose, ${2 m + 3} = r \\ 2 dm = dr \\ \int e ^{2 m + 3} dm = \frac{1}{2} \int e ^{r} dr \\ = \frac{1}{2} (e ^{r}) + C \\ = \frac{1}{2} (e ^{2 m + 3}) + C$

Question 17:

Obtain an integral (or anti – derivative) of the $\frac{u}{e ^{u ^{2}}}$

Suppose, u 2 = z

2u du = dz

$\int \frac{u}{e ^{u ^{2}}} du = \frac{1}{2} \int \frac{1}{e ^{z}} dz \\ = \frac{1}{2} \int e ^{- z} dz \\ = \frac{1}{2} \frac{e ^{- z}}{- 1} + C \\ = – \frac{1}{2} e ^{- u ^{2}} + C \\ = – \frac{1}{2 e ^{u ^{2}}} + C$

Question 18:

Obtain an integral (or anti – derivative) of the $\frac{e ^{tan ^{- 1} \Theta}}{1 + \Theta ^{2}}$

Suppose, $tan ^{- 1} \Theta = z \frac{1}{1 + \Theta ^{2}} d\Theta = dz \\ \int \frac{e ^{tan ^{- 1} \Theta}}{1 + \Theta ^{2}} d\Theta = \int e ^{z} dz \\ = e ^{z} + C \\ = e ^{tan ^{- 1} \Theta} + C$

Question 19:

Obtain an integral (or anti – derivative) of the $\frac{e ^{2u} – 1}{e ^{2u} + 1}$

$\frac{e ^{2u} – 1}{e ^{2u} + 1}$

Dividing the numerator and denominator by e u, we get

$\frac{\frac{e ^{2u} – 1}{e ^{u}}}{\frac{e ^{2u} + 1}{e ^{u}}} = \frac{e ^{u} – e ^{- u}}{e ^{u} + e ^{- u}} \\$

Suppose,

$e ^{u} + e ^{- u} = z \\ (e ^{u} – e ^{- u}) du = dz \\ \int \frac{e ^{2u} – 1}{e ^{2u} + 1} du = \int \frac{e ^{u} – e ^{- u}}{e ^{u} + e ^{- u}} du \\ = \int \frac{dz}{z} \\ = log \left | z \right | + C \\ = log \left | e ^{u} + e ^{- u} \right | + C$

Question 20:

Obtain an integral (or anti – derivative) of the $\frac{e ^{2u} – e ^{- 2u}}{e ^{2u} + e ^{- 2u}}$

Suppose, $e ^{2u} + e ^{- 2u} = z \\ (2 e ^{2u} – 2 e ^{- 2u}) du = dz \\ 2 (e ^{2u} – e ^{- 2u}) du = dz \\ \int \frac{e ^{2u} – e ^{- 2u}}{e ^{2u} + e ^{- 2u}} = \int \frac{dz}{2z} dz \\ = \frac{1}{2} \int \frac{1}{z} dz \\ = \frac{1}{2} log \left | z \right | + C \\ = \frac{1}{2} log \left | e ^{2u} + e ^{- 2u} \right | + C$

Question 21:

Obtain an integral (or anti – derivative) of the $tan ^{2} (2 \Theta – 3)$

$tan ^{2} (2 \Theta – 3) = sec ^{2} (2 \Theta – 3) – 1$

Suppose, $2 \Theta – 3 = z \\ 2 d\Theta = dz \\ \int tan ^{2} (2 \Theta – 3) d\Theta = \int [sec ^{2} (2 \Theta – 3) – 1] d\Theta \\ = \frac{1}{2} \int (sec ^{2} z) dz – \int 1 d \Theta \\ = \frac{1}{2} tan z – \Theta + C \\ = \frac{1}{2} tan (2 \Theta – 3) – \Theta + C$

Question 22:

Obtain an integral (or anti – derivative) of the $sec ^{2} (7 – 4 \theta)$

Suppose, $(7 – 4 \theta) = z \\ – 4\; d \theta = dz \\ \int sec ^{2} (7 – 4 \Theta) d\theta = – \frac{1}{4} \int sec ^{2} z dz \\ = – \frac{1}{4} (tan z) + C \\ = – \frac{1}{4} [tan (7 – 4 \theta)] + C$

Question 23:

Obtain an integral (or anti – derivative) of the $\frac{sin ^{- 1} \theta}{\sqrt{1 – \theta ^{2}}}$

Suppose, $sin ^{- 1} \theta = z \\ \frac{1}{\sqrt{1 – \theta ^{2}}} d\theta = dz \\ \int \frac{sin ^{- 1} \theta}{\sqrt{1 – \theta ^{2}}} d\theta = \int z dz \\ = \frac{z ^{2}}{2} + C \\ = \frac{(sin ^{- 1} \theta) ^{2}}{2} + C \\$

Question 24:

Obtain an integral (or anti – derivative) of the $\frac{2 cos\; \theta – 3 sin\; \theta}{6 cos\; \theta + 4 sin\; \theta}$

$\frac{2 cos\; \theta – 3 sin\; \theta}{6 cos\; \theta + 4 sin\; \theta} = \frac{2 cos\; \theta – 3 sin\; \theta}{2 (3 cos\; \theta + 2 sin\; \theta)}$

Suppose,

$3\; cos\; \theta + 2\; sin\; \theta = z\\ (- 3\; sin\; \theta + 2\; cos\; \theta) d\theta = dz \\ \int \frac{2 cos\; \theta – 3 sin\; \theta}{6 cos\; \theta + 4 sin\; \theta} d\theta = \int \frac{dz}{2z} \\ = \frac{1}{2} \frac{1}{z} dz \\ = \frac{1}{2} log \left | z \right | + C \\ = \frac{1}{2} log \left | 3\; cos\; \theta + 2\; sin\; \theta \right | + C \\$

Question 25:

Obtain an integral (or anti – derivative) of the $\frac{1}{cos ^{2} \theta (1 – tan \theta) ^{2}}$

$\frac{1}{cos ^{2} \theta (1 – tan \theta) ^{2}} = \frac{sec ^{2} \theta}{(1 – tan \theta) ^{2}} \\$

Suppose,

$(1 – tan \theta) = z \\ sec ^{2} \theta d\theta = dz \\ \int \frac{sec ^{2} \theta}{(1 – tan \theta) ^{2}} d\theta = \int -\frac{dz}{z ^{2}} \\ = – \int z ^{- 2} dz \\ = \frac{1}{z} + C \\ = \frac{1}{1 – tan \theta} + C$

Question 26:

Obtain an integral (or anti – derivative) of the $\frac{cos \sqrt{\theta }}{\sqrt{\theta }}$

Suppose, $\sqrt{\theta } = z \\ \frac{1}{2 \sqrt{\theta}} d\theta = dz \\ \int \frac{cos \sqrt{\theta }}{\sqrt{\theta }} = 2 \int cos\; z dz \\ = 2 sin\; z + C \\ = 2 sin\; \sqrt{\theta } + C$

Question 27:

Obtain an integral (or anti – derivative) of the $\sqrt{sin\; 2 \theta}\; cos\; 2 \theta$

Suppose, $sin\; 2 \theta = z \\ 2 cos\; 2 \theta d\theta = dz \\ \int \sqrt{sin\; 2 \theta}\; cos\; 2 \theta = \frac{1}{2} \int \sqrt{z} dz \\ = \frac{1}{2} (\frac{z ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ = \frac{1}{3} z ^{\frac{3}{2}} + C \\ = \frac{1}{3} (sin\; 2 \theta) ^{\frac{3}{2}} + C$

Question 28:

Obtain an integral (or anti – derivative) of the $\frac{cos\; \theta}{\sqrt{1 + sin\; \theta}}$

Suppose, $1 + sin\; \theta = z \\ cos\; \theta d\theta = dz \\ \int \frac{cos\; \theta}{\sqrt{1 + sin\; \theta}} d\theta = \int \frac{dz}{\sqrt{z}} \\ = \frac{z ^{\frac{1}{2}}}{\frac{1}{2}} + C \\ = 2 \sqrt{z} + C \\ = 2 \sqrt{1 + sin\; \theta} + C$

Question 29:

Obtain an integral (or anti – derivative) of the $cot\; \theta\; log\; sin\; \theta$

Suppose, $log\; sin\; \theta = z \\ \frac{1}{sin\; \theta }. cos\; \theta = dz \\ cot\; \theta\; d\theta = dz \\ \int cot\; \theta\; log\; sin\; \theta d\theta = \int z\; dz \\ = \frac{z ^{2}}{2} + C \\ = \frac{1}{2} (log\; sin\; \theta) ^{2} + C$

Question 30:

Obtain an integral (or anti – derivative) of the $\frac{sin\; \theta}{1 + cos\; \theta}$

Suppose,

$1 + cos\; \theta = z – sin\; \theta d\theta = dz \\ \int \frac{sin\; \theta}{1 + cos\; \theta} d\theta = \int – \frac{dz}{z } \\ = – \int \frac{dz}{z} dz \\ = – log \left | z \right | + C \\ = – log \left | 1 + cos\; \theta \right | + C$

Question 31:

Obtain an integral (or anti – derivative) of the $\frac{sin\; \theta}{(1 + cos\; \theta) ^{2}}$

Suppose, $1 + cos\; \theta = z – sin\; \theta d\theta = dz \\ \int \frac{sin\; \theta}{1 + cos\; \theta} d\theta = \int – \frac{dz}{z ^{2}} \\ = – \int \frac{dz}{z ^{2}} dz \\ = – \int z ^{- 2} dz \\ = \frac{1}{z} + C \\ = \frac{1}{1 + cos\; \theta} + C$

Question 32:

Obtain an integral (or anti – derivative) of the $\frac{1}{1 + cot\; \theta}$

Suppose, I = $\int \frac{1}{1 + cot\; \theta} d\theta \\ = \int \frac{1}{1 + \frac{cos\; \theta}{sin\; \theta}} d\theta \\ = \int \frac{sin\; \theta}{sin\; \theta + cos\; \theta} d\theta \\ = \frac{1}{2} \int \frac{2 sin\; \theta}{sin\; \theta + cos\; \theta} d\theta \\ = \frac{1}{2} \int \frac{(sin\; \theta + cos\; \theta) + (sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\ = \frac{1}{2} \int 1 d\theta + \frac{1}{2} \int \frac{(sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\ = \frac{1}{2} \theta + \frac{1}{2} \int \frac{(sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\$ $Suppose,\; (sin\; \theta + cos\; \theta) = z \\ = (cos\; \theta – sin\; \theta) d\theta = dz \\ I = \frac{\theta }{2} + \frac{1}{2} log \left | z \right | + C \\ = \frac{\theta }{2} – \frac{1}{2} log \left | (sin\; \theta + cos\; \theta) \right | + C \\$

Question 33:

Obtain an integral (or anti – derivative) of the $\frac{1}{1 – tan \theta}$

Suppose,

$\int \frac{1}{1 – tan\; \theta} d\theta \\ = \int \frac{1}{1 – \frac{sin\; \theta}{cos\; \theta}} d\theta \\ = \int \frac{cos\; \theta}{cos\; \theta – sin\; \theta} d\theta \\ = \frac{1}{2} \int \frac{2 cos\; \theta}{cos\; \theta – sin\; \theta} d\theta \\ = \frac{1}{2} \int \frac{(cos\; \theta – sin\; \theta ) + (cos\; \theta + sin\; \theta)}{(cos\; \theta – sin\; \theta)} d\theta \\ = \frac{1}{2} \int 1 d\theta + \frac{1}{2} \int \frac{(cos\; \theta + sin\; \theta)}{(cos\; \theta – sin\; \theta)} d\theta \\ = \frac{1}{2} \theta + \frac{1}{2} \int \frac{(cos\; \theta + sin\; \theta)}{(cos\; \theta – sin\; \theta)} d\theta \\$ $Suppose,\; (cos\; \theta – sin\; \theta) = z \\ = (- sin\; \theta – cos\; \theta ) d\theta = dz \\ I = \frac{\theta }{2} – \frac{1}{2} log \left | z \right | + C \\ = \frac{\theta }{2} – \frac{1}{2} log \left | (cos\; \theta – sin\; \theta) \right | + C \\$

Question 34:

Obtain an integral (or anti – derivative) of the $\frac{\sqrt{tan \theta}}{sin\; \theta\; cos\; \theta}$

Suppose, $Suppose, I = \frac{\sqrt{tan \theta}}{sin\; \theta\; cos\; \theta} d\theta \\ = \frac{\sqrt{tan \theta} \times cos\; \theta}{sin\; \theta\; cos\; \theta \times cos\; \theta} d\theta \\ = \int \frac{\sqrt{tan \theta}}{tan\; \theta\; cos\; ^{2} \theta} d\theta \\ = \int \frac{sec\; ^{2} \theta}{\sqrt{tan\; \theta}} d\theta \\ Suppose, tan\; \theta = z \\ sec\; ^{2} \theta d\theta = dz \\ I = \int \frac{dz}{\sqrt{z}} \\ = 2 \sqrt{z} + C \\ = 2 \sqrt{tan\; \theta} + C$

Question 35:

Obtain an integral (or anti – derivative) of the $\frac{(1 + log u) ^{2}}{u}$

Suppose, $Suppose, 1 + log u = z \\ \frac{1}{u} du = dz \\ \int \frac{(1 + log u) ^{2}}{u} du = \int z ^{2}\; dz \\ = \frac{z ^{3}}{3} + C \\ = \frac{(1 + log u) ^{3}}{3} + C$

Question 36:

Obtain an integral (or anti – derivative) of the $\frac{(u + 1)(u + log u) ^{2}}{u}$

$\frac{(u + 1)(u + log u) ^{2}}{u} = \frac{(u + 1)}{u} (u + log u) ^{2} = (1 + \frac{1}{u}) (u + log u) ^{2} \\ Suppose,\; (u + log u) = z \\ (1 + \frac{1}{u}) du = dz \\ \int (1 + \frac{1}{u}) (u + log\; u) ^{2} du = \int z ^{2} dz \\ = \frac{z ^{3}}{3} + C \\ = \frac{1}{3} (u + log u) ^{3} + C$

Question 37:

Obtain an integral (or anti – derivative) of the $\frac{u ^{3}\; sin\; (tan ^{- 1} u ^{4})}{1 + u ^{8}}$

$Suppose,\; u ^{4} = z \\ 4 u ^{3} du = dz \\ \int \frac{u ^{3}\; sin\; (tan ^{- 1} u ^{4})}{1 + u ^{8}} du = \frac{1}{4} \int \frac{sin\; (tan ^{- 1} z)}{1 + z ^{2}} \;dz …. (1) \\ Suppose,\; tan ^{- 1} z = s \\ \frac{1}{1 + z ^{2}} dz = ds \\ From\; (1),\; we\; get\; \\ \int \frac{u ^{3}\; sin\; (tan ^{- 1} u ^{4})}{1 + u ^{8}} du = \frac{1}{4} \int sin\; s\; ds \\ = \frac{1}{4} (- cos s) + c \\ = – \frac{1}{4} cos (tan ^{- 1} z) + C \\ = – \frac{1}{4} cos (tan ^{- 1} u ^{4}) + C \\$

Question 38:

Which of the following below is the answer for $\int \frac{10 u ^{9} + 10 ^{u} log_{e} 10}{u ^{10} + 10 ^{u}} du$ :

$(a) 10 ^{u} – u ^{10} + C \\ (b) 10 ^{u} + u ^{10} + C \\ (c) (10 ^{u} – u ^{10}) ^{- 1} + C \\ (d) log (10 ^{u} + u ^{10}) + C \\$

$u ^{10} + 10 ^{u} = z \\ (10 u ^{9} + 10 ^{u} log_{e} 10) du = dz \\ \int \frac{10 u ^{9} + 10 ^{u} log_{e} 10}{u ^{10} + 10 ^{u}} du = \int \frac{dz}{z} \\ = log z + C \\ = log (u ^{10} + 10 ^{u}) + C \\ Therefore,\; D\; is\; the\; correct\; answer$

Question 39:

Which of the following below is the answer for $\int \frac{du}{sin ^{2} \;u\; cos ^{2} \;u}$ $(a) tan\; u + cot\; u + C \\ (b) tan\; u – cot\; u + C \\ (c) tan\; u \; cot\; u + C \\ (d) tan\; u – cot\; 2u + C \\$

$I = \int \frac{du}{sin ^{2} \;u\; cos ^{2} \;u} \\ = \int \frac{1}{sin ^{2} \;u\; cos ^{2} \;u} du \\ = \int \frac{sin ^{2} \;u\; + cos ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u} du \\ = \int \frac{sin ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u} du + \int \frac{cos ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u} du \\ = \int sec ^{2} \;u du + \int cosec ^{2} \;u du \\ = tan\; u – cot\; u + C \\$

Therefore, B is the correct answer

Exercise 7.3

Question 1:

Obtain an integral (or anti – derivative) of the sin 2 (2u + 5)

$sin ^{2} (2u + 5) = \frac{1 – cos\; 2 (2u + 5)}{2} = \frac{1 – cos\; (4u + 10)}{2} \\ \int sin ^{2} (2u + 5) du = \int \frac{1 – cos\; (4u + 10)}{2} du \\ = \frac{1}{2} \int 1 du – \frac{1}{2} \int cos\; (4u + 10) du \\ = \frac{1}{2} u – \frac{1}{2} \frac{sin (4u + 10)}{4} + C \\ = \frac{1}{2} u – \frac{1}{8} [sin (4u + 10)] + C$

Question 2:

Obtain an integral (or anti – derivative) of the sin 3u. cos 4u

As we know, $sin\; C cos\; D = \frac{1}{2} [sin\; (C + D) + sin\; (C – D)] \\ \int sin\; 3u. cos\; 4u\; du = \int \frac{1}{2} [sin\; (3u + 4u) + sin\; (3u – 4u)] \\ = \int \frac{1}{2} [sin\; (7u) + sin\; (- u)] du \\ = \int \frac{1}{2} [sin\; (7u) – sin\; (u)] du \\ = \frac{1}{2} \int sin\; (7u)\; du – \frac{1}{2} \int sin\; (u)\; du \\ = \frac{1}{2} (\frac{- cos\; 7u}{7}) – \frac{1}{2} (- cos\; u) + C \\ = \frac{- cos\; 7u}{14} + \frac{cos\; u}{2} + C$

Question 3:

Obtain an integral (or anti – derivative) of the cos 2u cos 4u cos 6u

As we know, $cos\; C\; cos\; D = \frac{1}{2} [cos\; (C + D) + cos\; (C – D)] \\ \int cos\; 2u\; (cos\; 4u\; cos\; 6u) du = \int cos\; 2u\; [\frac{1}{2} (cos\; (4u + 6u) + cos\; (4u – 6u))] du \\ \frac{1}{2} \int [cos\; 2u\; (cos\; (10 u) + cos\; (- 2u))] du \\ \frac{1}{2} \int [cos\; 2u\; cos\; 10 u + cos\; 2u\; cos\; (- 2u)] du \\ \frac{1}{2} \int [cos\; 2u\; cos\; 10 u + cos ^{2} \;2u]\; du \\ \frac{1}{2} \int [\frac{1}{2}\; (cos\; (2u + 10u) + cos\; (2u – 10u)) + (\frac{1 + cos\; 4u}{2})]\; du \\ \frac{1}{4} \int (cos\; 12u + cos\; 8u + 1 + cos\; 4u)\; du \\ \frac{1}{4} [\frac{sin 12u}{12} + \frac{sin 8u}{8} + u + \frac{sin 4u}{4}] + C \\$

Question 4:

Obtain an integral (or anti – derivative) of the $sin ^{3} (2u + 1)$

$I = \int sin ^{3} (2u + 1) du \\ \int sin ^{3} (2u + 1) du = \int sin ^{2} \;(2u + 1). sin \;(2u + 1)\; du \\ Supose,\; cos\; (2u + 1) = z \\ – 2 sin (2u + 1)\; du = dz \\ sin (2u + 1)\; du = \frac{- dz}{2} \\ I = \frac{- 1}{2} \int (1 – z^{2}) dz \\ = \frac{- 1}{2} \left \{ z – \frac{z ^{3}}{3} \right \} + C \\ = \frac{- 1}{2} \left \{cos\; (2u + 1) – \frac{cos ^{3} \;(2u + 1)}{3} \right \} + C \\ = \frac{- cos\; (2u + 1)}{2} + \frac{cos ^{3} \;(2u + 1)}{6} + C \\$

Question 5:

Obtain an integral (or anti – derivative) of the $sin^{3} \;u\; cos^{3} \;u$

$I = \int sin^{3} \;u\; cos^{3} \;u\; du \\ = \int cos^{3} \;u\;. sin^{2} \;u\; sin\; u\; du \\ = \int cos^{3} \;u\; (1 – cos^{2} \;u). sin\; u\; du \\ Suppose,\; cos\; u = z \\ – sin\; u\; du = dz \\ I = – \int z^{3} (1 – z ^{2}) dz \\ = – \int (z^{3} – z ^{5}) dz \\ = – \left \{ \frac{z ^{4}}{4} – \frac{z ^{6}}{6} \right\} + C \\ = – \left \{ \frac{cos ^{4}}{4} – \frac{cos ^{6}}{6} \right\} + C \\ = \frac{cos ^{6}}{6} – \frac{cos ^{4}}{4} + C$

Question 6:

Obtain an integral (or anti – derivative) of the sin u sin 2u sin 3u

$sin\; C\; sin\; D = \frac{1}{2} [cos\; (C – D) – cos\; (C + D)] \\ \int sin\; u\; sin\; 2u\; sin\; 3u\; du = \int sin\; u\;. \frac{1}{2} \left \{ cos\; (2u – 3u) – cos\; (2u + 3u)\right \}\; du \\ = \frac{1}{2} \int (sin\; u\; cos\; (- u) – sin\; u\; cos\; 5u) \; du \\ = \frac{1}{2} \int (sin\; u\; cos\; u – sin\; u\; cos\; 5u) \; du \\ = \frac{1}{2} \int \frac{(2 sin\; u\; cos\; u)}{2} \;du – \frac{1}{2} \int sin\; u\; cos\; 5u \;du \\ = \frac{1}{2} \int \frac{(sin\; 2u)}{2} \;du – \frac{1}{2} \int sin\; u\; cos\; 5u \;du \\$ $= \frac{1}{4} \left [ \frac{- cos\; 2u}{2} \right ] – \frac{1}{2} \int \left \{ \frac{1}{2} (sin\; (u + 5u) + sin\; (u – 5u)) \right \} du \\ = \frac{- cos\; 2u}{8} – \frac{1}{4} \int \left \{(sin\; (6u) + sin\; (- 4u)) \right \} du \\ = \frac{- cos\; 2u}{8} – \frac{1}{4} \left [ \frac{- cos\; 6u}{6} + \frac{cos\; 4u}{4} \right ] + C \\ = \frac{- cos\; 2u}{8} – \frac{1}{8} \left [ \frac{- cos\; 6u}{3} + \frac{cos\; 4u}{2} \right ] + C \\ = \frac{1}{8} \left [ \frac{- cos\; 6u}{3} – \frac{cos\; 4u}{2} – cos\; 2u \right ] + C$

Question 7:

Obtain an integral (or anti – derivative) of the sin 4u sin 8u

As we know, $sin\; C\; sin\; D = \frac{1}{2} [cos\; (C – D) – cos\; (C + D)] \\ \int sin\; 4u\; sin\; 8u\; du = \int \frac{1}{2} [cos\; (4u – 8u) – cos\; (4u + 8u)]\; du \\ = \frac{1}{2} \int (cos\; (- 4u) – cos\; 12u)\; du \\ = \frac{1}{2} \int (cos\; 4u – cos\; 12u)\; du \\ = \frac{1}{2} \left [ \frac{sin\; 4u}{4} – \frac{sin\; 12u}{12} \right ]$

Question 8:

Obtain an integral (or anti – derivative) of the $\frac{1 – cos\; u}{1 + cos\; u}$

$\frac{1 – cos\; u}{1 + cos\; u} = \frac{2 sin^{2}\; \frac{u}{2}}{2 cos^{2}\; \frac{u}{2}} \\ = tan^{2}\; \frac{u}{2} \\ = (sec^{2}\; \frac{u}{2} – 1) \\ \int \frac{1 – cos\; u}{1 + cos\; u} du = \int (sec^{2}\; \frac{u}{2} – 1) \;du \\ = \left [ \frac{tan\; \frac{u}{2}}{\frac{1}{2}} – u \right ] + C \\ = 2 tan\; \frac{u}{2} – u + C$

Question 10:

Obtain an integral (or anti – derivative) of the sin 4 u.

$sin ^{4} \;u = sin ^{2} \;u \times sin ^{2} \;u \\ = (\frac{1 – cos \;2u}{2}) (\frac{1 – cos \;2u}{2}) \\ = \frac{1}{4} (1 – cos \;2u)^{2} \\ = \frac{1}{4} (1 + cos^{2} \;2u – 2\; cos \;2u) \\ = \frac{1}{4} \left [ 1 + (\frac{1 + cos\; 4u}{2}) – 2 cos\; 2u \right ] \\ = \frac{1}{4} [1 + \frac{1}{2} + \frac{1}{2}\; cos\; 4u – 2 cos\; 2u] \\ = \frac{1}{4} \left [ \frac{3}{2} + \frac{1}{2}\; cos\; 4u – 2 cos\; 2u\right ] \\$ $\int sin ^{4} \;u\; du = \frac{1}{4}\; \int \left [ \frac{3}{2} + \frac{1}{2}\; cos\; 4u – 2 cos\; 2u\right ] du \\ = \frac{1}{4}\; \left [ \frac{3}{2} u + \frac{1}{2}\; (\frac{sin\; 4u}{4}) – sin\; 2u \right ] + C \\ = \frac{3}{8} u + (\frac{sin\; 4u}{32}) – \frac{sin\; 2u}{4} + C$

Question 11:

Obtain an integral (or anti – derivative) of the cos 4 2u

$cos ^{4} \;2u = (sin ^{2} \;2u) ^{2} \\ = (\frac{1 + cos \;4u}{2}) ^{2} \\ = \frac{1}{4} (1 + cos^{2} \;4u + 2\; cos \;4u) \\ = \frac{1}{4} \left [ 1 + (\frac{1 + cos\; 8u}{2}) + 2 cos\; 4u \right ] \\ = \frac{1}{4} [1 + \frac{1}{2} + \frac{1}{2}\; cos\; 8u + 2 cos\; 4u] \\ = \frac{1}{4} \left [ \frac{3}{2} + \frac{1}{2}\; cos\; 8u + 2 cos\; 4u\right ] \\$ $\int cos ^{4} \;u\; du = \frac{1}{4}\; \int \left [ \frac{3}{2} + \frac{1}{2}\; cos\; 8u + 2 cos\; 4u\right ] du \\ = \frac{1}{4}\; \left [ \frac{3}{2} u + \frac{1}{2}\; (\frac{sin\; 8u}{4}) + sin\; 4u \right ] + C \\ = \frac{3}{8} u + (\frac{sin\; 8u}{64}) + \frac{sin\; 4u}{8} + C$

Question 12:

Obtain an integral (or anti – derivative) of the $\frac{sin ^{2} \;u}{1 + cos\; u}$

$\frac{sin ^{2} \;u}{1 + cos\; u} = \frac{(2 sin\; \frac{u}{2}\; cos\; \frac{u}{2}) ^{2}}{2\; cos^{2} \frac{u}{2}} \;\;\;[Since\; sin\; u = 2\; sin\; \frac{u}{2}\; cos\; \frac{u}{2}; \; cos\; u = 2\; cos^{2} \frac{u}{2} – 1] \\ = \frac{4 sin ^{2} \;\frac{u}{2}\; cos ^{2} \;\frac{u}{2}}{2\; cos^{2} \frac{u}{2}} \\ = 2 sin ^{2} \;\frac{u}{2} \\ = 1 – cos\; u \\ \int \frac{sin ^{2} \;u}{1 + cos\; u} du = \int (1 – cos\; u) \; du \\ = u – sin\; u + C$

Question 13:

Obtain an integral (or anti – derivative) of the $\frac{cos\; 2u – cos\; 2a}{cos\; u – cos\; a}$

$\frac{cos\; 2u – cos\; 2a}{cos\; u – cos\; a} = \frac{- 2\; sin \;\frac{2u + 2a}{2}\; sin \;\frac{2u – 2a}{2}}{- 2\; sin \;\frac{u + a}{2}\; sin \;\frac{u – a}{2}} \; \; \left [Since,\; cos\; A – cos\; B = – 2\; sin\; \frac{A + B}{2}\; sin\; \frac{A – B}{2} \right ] \\ = \frac{sin\; (u + a)\; sin\; (u – a)}{ sin \;\frac{u + a}{2}\; sin \;\frac{u – a}{2}} \\ = \frac{\left [ 2 sin\; \frac{u + a}{2}\; cos\; \frac{u + a}{2} \right ] \left [ 2 sin\; \frac{u – a}{2}\; cos\; \frac{u – a}{2} \right ]}{sin\; \frac{u + a}{2}\; sin\; \frac{u – a}{2}} \\$ $= 4 cos\; \frac{u + a}{2}\; cos\; \frac{u – a}{2} \\ = 2 \left [ cos\; \left ( \frac{u + a}{2} + \frac{u – a}{2} \right ) + cos\; \left ( \frac{u + a}{2} – \frac{u – a}{2} \right ) \right ] \\ = 2 [cos\; (u) + cos\; a] \\ = 2 cos\; u + 2 cos\; a$

Question 14:

Obtain an integral (or anti – derivative) of the $\frac{cos\; u – sin\; u}{1 + sin\; 2u}$

$\frac{cos\; u – sin\; u}{1 + sin\; 2u} = \frac{cos\; u – sin\; u}{(sin ^{2} \;u + cos ^{2} \;u) + 2 sin\; u\; cos\; u} \\ Since,\; sin ^{2} \;u + cos ^{2} \;u = 1; sin\; 2u = 2 sin\; u\; cos\; u \\ = \frac{cos\; u – sin\; u}{(sin\; u + cos\; u) ^{2}} \\ Suppose,\; sin\; u + cos\; u = z \\ (cos\; u – sin\; u)\; du = dz \\ \int \frac{cos\; u – sin\; u}{1 + sin\; 2u}\; du = \int \frac{cos\; u – sin\; u}{(sin\; u + cos\; u) ^{2}} du \\$ $= \frac{dz}{z ^{2}} \\ = \int z ^{- 2} dz \\ = – z ^{- 1} + C \\ = – \frac{1}{z} + C \\ = \frac{- 1}{sin\; u + cos\; u} + C$

Question 15:

Obtain an integral (or anti – derivative) of the $tan ^{3}\; 2u\; sec\; 2u$

$tan ^{3}\; 2u\; sec\; 2u = tan ^{2}\; 2u\; tan\; 2u\; sec\; 2u \\ = (sec ^{2} \;2u – 1) tan\; 2u\; sec\; 2u \\ = sec ^{2} \;2u\; tan\; 2u\; sec\; 2u – tan\; 2u\; sec\; 2u \\ = \int tan ^{3}\; 2u\; sec\; 2u du = \int sec ^{2} \;2u\; tan\; 2u\; sec\; 2u\; du – \int tan\; 2u\; sec\; 2u \;du \\ = \int sec ^{2} \;2u\; tan\; 2u\; sec\; 2u\; du – \frac{sec\; 2u}{2} + C \\$ $Suppose,\; sec\; 2u = z \\ 2 sec\; 2u\; tan\; 2u du = dz \\ \int tan ^{3}\; 2u\; sec\; 2u = \frac{1}{2} z ^{2} dz – \frac{sec\; 2u}{2} + C \\ = \frac{z ^{3}}{6} – \frac{sec\; 2u}{2} + C \\ = \frac{(sec\; 2u) ^{3}}{6} – \frac{sec\; 2u}{2} + C$

Question 16:

Obtain an integral (or anti – derivative) of the $tan ^{4} \;u$

$tan ^{4} \;u = tan ^{2} \;u\; \times tan ^{2} \;u \\ = (sec ^{2} \;u\; – 1) tan ^{2} \;u \\ = sec ^{2} \;u\; tan ^{2} \;u – tan ^{2} \;u \\ = sec ^{2} \;u\; tan ^{2} \;u – (sec ^{2} \;u\; – 1) \\ = sec ^{2} \;u\; tan ^{2} \;u – sec ^{2} \;u\; + 1 \\$ $\int tan ^{4} \;u\; du = \int sec ^{2} \;u\; tan ^{2} \;u\; du – \int sec ^{2} \;u\; du + \int 1\; du \\ = \int sec ^{2} \;u\; tan ^{2} \;u\; du – tan\; u + u + C \;\; …… (1) \\ Now,\; \int sec ^{2} \;u\; tan ^{2} \;u\; du \\ Suppose,\; tan\; u = z \\ sec ^{2} \;u\; du = dz \\ \int sec ^{2} \;u\; tan ^{2} \;u\; du = \int z ^{2}\; dz \\ = \frac{z ^{3}}{3} + C \\ = \frac{tan ^{3} \;u}{3} \\ Therefore, from equation (1) is\\ \int tan ^{4} \;u\; du = \frac{tan ^{3} \;u}{3} – tan\; u + u + C$

Question 17:

Obtain an integral (or anti – derivative) of the $\frac{sin ^{3} \;u + cos ^{3} \;u}{sin ^{2} \;u\; cos ^{2} \;u}$

$\frac{sin ^{3} \;u + cos ^{3} \;u}{sin ^{2} \;u\; cos ^{2} \;u} = \frac{sin ^{3} \;u}{sin ^{2} \;u\; cos ^{2} \;u} + \frac{cos ^{3} \;u}{sin ^{2} \;u\; cos ^{2} \;u} = \frac{sin \;u}{cos ^{2} \;u} + \frac{cos \;u}{sin ^{2} \;u} \\$ $= tan\; u\; sec\; u + cot\; u\; cosec\; u \\ Therefore,\; \int \frac{sin ^{3} \;u + cos ^{3} \;u}{sin ^{2} \;u\; cos ^{2} \;u}\; du = \int (tan\; u\; sec\; u + cot\; u\; cosec\; u)\; du \\ = sec\; u – cosec\; u + C \\$

Question 18:

Obtain an integral (or anti – derivative) of the $\frac{cos\; 2u + 2 sin ^{2} \;u}{cos ^{2} \;u} \\$

$\frac{cos\; 2u + 2 sin ^{2} \;u}{cos ^{2} \;u} \\ \frac{cos\; 2u + (1 – cos\; 2u)}{cos ^{2} \;u}\; \; [Since,\; cos\; 2u = 1 – 2 sin ^{2} \;u] \\ = \frac{1}{cos ^{2} \;u} \\ = sec ^{2} \;u \\ \int \frac{cos\; 2u + 2 sin ^{2} \;u}{cos ^{2} \;u}\; du = \int sec ^{2} \;u\; du = tan\; u + C$

Question 19:

Obtain an integral (or anti – derivative) of the $\frac{1}{sin\; u\; cos^{2} \;u}$

$\frac{1}{sin\; u\; cos^{2} \;u} = \frac{sin^{2} \;u + cos^{2} \;u}{sin\; u\; cos^{2} \;u} \\ = \frac{sin\; u\;}{cos^{2} \;u} + \frac{1}{sin\; u\; cos\; u} \\ = tan\; u\; sec^{2} \;u + \frac{1}{sin\; u\; cos\; u} \times \frac{cos ^{2} \;u}{cos ^{2} \;u} \\ = tan\; u\; sec^{2} \;u + \frac{sec ^{2} \;u}{tan\; u}$ $\int \frac{1}{sin\; u\; cos^{2} \;u} du = \int tan\; u\; sec^{2} \;u\; du + \int \frac{sec ^{2} \;u}{tan\; u}\; du \\$ $Suppose,\; tan\; u = z \\ sec ^{2} \;u\; du = dz \\ \int \frac{1}{sin\; u\; cos^{2} \;u}\; du = \int z\; dz + \int \frac{1}{z}\; dz\; \\ = \frac{z ^{2}}{2} + log \left | z \right | + C \\ = \frac{1}{2} tan^{2} \;u + log \left | tan\; u \right | + C$

Question 20:

Obtain an integral (or anti – derivative) of the $\frac{cos\; 2u}{(cos\; u + sin\; u) ^{2}}$

$\frac{cos\; 2u}{(cos\; u + sin\; u) ^{2}} = \frac{cos\; 2u}{cos ^{2} \;u + sin^{2} \;u + 2 sin\; u\; cos\; u} = \frac{cos\; 2u}{1 + sin\; 2u} \\ \int \frac{cos\; 2u}{(cos\; u + sin\; u) ^{2}}\; du = \int \frac{cos\; 2u}{(1 + sin\; 2u)}\; du \\ Suppose,\; 1 + sin\; 2u = Z \\ 2 cos\; 2u\; du = dz \\ \int \frac{cos\; 2u}{(cos\; u + sin\; u) ^{2}}\; du = \frac{1}{2} \int \frac{1}{z}\; dz \\ = \frac{1}{2} log \left | z \right | + C \\ = \frac{1}{2} log \left | 1 + sin\; 2u \right | + C \\ = \frac{1}{2} log \left | (cos\; u + sin\; u) ^{2} \right | + C \\ = \frac{1}{2} log \left | cos\; u + sin\; u \right | + C$

Question 21:

Obtain an integral (or anti – derivative) of the $sin ^{- 1} (cos\; u)$

$sin ^{- 1} (cos\; u) \\ Suppose,\; cos\; x = z \\ Then, sin\; u = \sqrt{1 – u ^{2}} \\ (- sin\; u)\; du = dz \\ du = \frac{- dz}{\sqrt{1 – z ^{2}}} \\ Therefore,\; \int sin ^{- 1} (cos\; u)\; du = \int sin ^{- 1}\; z (- \frac{dz}{\sqrt{1 – z ^{2}}})\; dz \\ Suppose,\; sin ^{- 1} \;z = p \\ \frac{1}{\sqrt{1 – z ^{2}}} \;dz = dp \\$ $Therefore,\; \int sin ^{- 1} (cos\; u)\; du = – \int p\; dp \\ = – \frac{p ^{2}}{2} + C \\ = – \frac{(sin ^{- 1}\; z) ^{2}}{2} + C \\ = – \frac{(sin ^{- 1}\; (cos\; u)) ^{2}}{2} + C ….. (1) \\ As\; we\; know, \\ sin ^{- 1} \;u + cos ^{- 1} \;u\; = \frac{\pi }{2} \\ Therefore,\; sin ^{- 1} (cos\; u) = \frac{\pi }{2} – cos ^{- 1} (cos\; u) = (\frac{\pi }{2} – u) \\ On\; substituting\; in\; equation\; (1), we\; get,\;$ $\int sin ^{- 1} (cos\; u)\; du = \frac{- \left [ \frac{\pi}{2} – u \right ] ^{2}}{2} + C \\ = – \frac{1}{2}\; \left ( (\frac{\pi }{2}) ^{2} + u^{2} – \pi u \right ) + C \\ = – \frac{(\pi )^{2}}{8} – \frac{u ^{2}}{2} + \frac{1}{2} \pi u + C \\ = \frac{\pi u}{2} – \frac{u ^{2}}{2} + \left ( c – \frac{(\pi )^{2}}{8} \right ) = \frac{\pi u}{2} – \frac{u ^{2}}{2} + C_{1}$

Question 22:

Obtain an integral (or anti – derivative) of the $\frac{1}{cos\; (u – m)\; cos\; (u – n)}$

$\frac{1}{cos\; (u – m)\; cos\; (u – n)} = \frac{1}{sin\; (m – n)} \left [ \frac{sin\; (m – n)}{cos\; (u – m)\; cos\; (u – n)} \right ] \\ = \frac{1}{sin\; (m – n)} \left [ \frac{sin\; [(u – n) – (u – m)]}{cos\; (u – m)\; cos\; (u – n)} \right ] \\ = \frac{1}{sin\; (m – n)} \frac{sin\; (u – n)\; cos\; (u – m) – cos\; (u – n)\; sin\; (u – m)}{cos\; (u – m)\; cos\; (u – n)} \\ = \frac{1}{sin\; (m – n)} [tan\; (u – n) – tan\; (u – m)]$ $\int \frac{1}{cos\; (u – m)\; cos\; (u – n)} du = \frac{1}{sin\; (m – n)} \int [tan\; (u – n) – tan\; (u – m)] du \\ = \frac{1}{sin\; (m – n)} [- log \left | cos\; (u – n) \right | + log \left | cos\; (u – m) \right |] \\ = \frac{1}{sin\; (m – n)} log\; \left [ \frac{cos\; (u – m)}{cos\; (u – n)} \right ] + C \\$

Question 23:

Which of the following below is the answer for $\frac{sin ^{2} \;u – cos ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u}$

(a) tan u + cot u + C

(b) tan u + cosec u + C

(c) – tan u + cot u + C

(d) tan u + sec u + C

$\int \frac{sin ^{2} \;u – cos ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u}\; du \\ = \int \left ( \frac{sin ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u} – \frac{cos ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u} \right )\; du \\ = \int (sec ^{2}\; u\; – cosec ^{2}\; u)\; du \\ = tan\; u + cot\; u + C$

Thus, (a) is the correct answer.

Question 24:

Which of the following below is the answer for $\int \frac{e ^{u} (1 + u)}{cos ^{2}\; (e ^{u} u)}\; du$

(a) – cot (e uu) + C

(b) tan (e uu) + C

(c) tan (eu) + C

(d) cot (eu) + C

$\int \frac{e ^{u} (1 + u)}{cos ^{2}\; (e ^{u} u)}\; du \\ Suppose,\; e ^{u} u = z \\ (e ^{u}. u + e ^{u}. 1) du = \int \frac{dz}{cos ^{2}\; z} \\ = \int sec ^{2} \;z dz \\ = tan\; z + C \\ = tan (e ^{u}. u) + C$

Thus, (b) is the correct answer.

Exercise 7.4

Question 1:

Obtain an integral (or anti – derivative) of the $\frac{3 u^{2}}{u^{6} + 1}$

$Suppose,\; u ^{3} = z \\ 3 u ^{2} du = dz \\ \int \frac{3 u^{2}}{u^{6} + 1} du = \int \frac{dz}{z ^{2} + 1} \\ = tan ^{- 1}\; z + C \\ = tan ^{- 1}\; u ^{3} + C$

Question 2:

Obtain an integral (or anti – derivative) of the $\frac{1}{\sqrt{1 + 4 u ^{2}}}$

$Suppose,\; 2 u = z \\ 2\; du = dz \\ \int \frac{1}{\sqrt{1 + 4 u ^{2}}}\; du = \frac{1}{2} \int \frac{dz}{\sqrt{1 + z ^{2}}} \\ = \frac{1}{2} [log \left | z + \sqrt{1 + z ^{2}} \right |] + C \\ = \frac{1}{2} [log \left | 2 u + \sqrt{1 + 4 u^{2}} \right |] + C$

Question 3:

Obtain an integral (or anti – derivative) of the $\frac{1}{\sqrt{(2 – u)^{2} + 1}}$

$Suppose,\; 2 – u = z \\ – du = dz \\ \int \frac{1}{\sqrt{(2 – u)^{2} + 1}}\; du = – \int \frac{1}{\sqrt{z ^{2} + 1}} \;dz \\ = – [log \left | z + \sqrt{z ^{2} + 1} \right |] + C \\ = – [log \left | 2 – u + \sqrt{(2 – u) ^{2} + 1} \right |] + C \\ = log \left | \frac{1}{(2 – u) + \sqrt{u ^{2} – 4u + 5}} \right | + C$

Question 4:

Obtain an integral (or anti – derivative) of the $\frac{1}{\sqrt{9 – 25 u^{2}}}$

Suppose, 5u = z

5 du = dz

$\int \frac{1}{\sqrt{9 – 25 u^{2}}}\; du = \frac{1}{5} \int \frac{1}{9 – z ^{2}}\; dz \\ = \frac{1}{5} \int \frac{1}{\sqrt{3 ^{2} – z ^{2}}}\; dz \\ = \frac{1}{5} sin ^{- 1} \;\frac{z}{3} + C \\ = \frac{1}{5} sin ^{- 1} \;\frac{5 u}{3} + C \\$

Question 5:

Obtain an integral (or anti – derivative) of the $\frac{3 u}{1 + 2 u ^{4}}$

$Suppose,\; \sqrt{2} u^{2} = z \\ 2 \sqrt{2}\; u\; du = dz \\ \int \frac{3 u}{1 + 2 u ^{4}}\; du = \frac{3}{2 \sqrt{2}} \int \frac{dz}{1 + z ^{2}} dz \\ = \frac{3}{2 \sqrt{2}} [tan ^{- 1} z] + C \\ = \frac{3}{2 \sqrt{2}} [tan ^{- 1} (\sqrt{2} u^{2})] + C$

Question 6:

Obtain an integral (or anti – derivative) of the $\frac{u ^{2}}{1 – u ^{6}}$

Suppose, u3 = z

3 u2 du = dz

$\int \frac{u ^{2}}{1 – u ^{6}} du = \frac{1}{3} \int \frac{dz}{1 – z^{2}} \\ = \frac{1}{3} \left [ \frac{1}{2} log \left |\frac{1 + z}{1 – z} \right |\right ] + C \\ = \frac{1}{6} log \left | \frac{1 + u^{3}}{1 – u^{3}} \right | + C$

Question 7:

Obtain an integral (or anti – derivative) of the $\frac{u – 1}{u ^{2} – 1}$

$\int \frac{u – 1}{\sqrt{u ^{2} – 1}} du = \int \frac{u}{\sqrt{u ^{2} – 1}}\; du – \frac{1}{\sqrt{u ^{2} – 1}}\; du \\ For, \int \frac{u}{\sqrt{u ^{2} – 1}}\; du \\$ $Suppose\; u ^{2} – 1 = z \\ 2u\; du = dz \\ Therefore,\; \int \frac{u}{\sqrt{u ^{2} – 1}}\; du = \frac{1}{2} \int \frac{dz}{\sqrt{z}} \\ = \frac{1}{2} \int z ^{- \frac{1}{2}} dz \\ = \frac{1}{2} [2 z^{\frac{1}{2}}] + C \\ = \sqrt{z} + C \\ = \sqrt{u ^{2} – 1} + C$ $From\; the\; above\; equation\; we\; get \\ \int \frac{u – 1}{\sqrt{u ^{2} – 1}} du = \int \frac{u}{\sqrt{u ^{2} – 1}}\; du – \frac{1}{\sqrt{u ^{2} – 1}}\; du \\ = \sqrt{u ^{2} – 1} + C – log \left | u + \sqrt{u ^{2} – 1} \right | + C_{1} \\ = \sqrt{u ^{2} – 1} – log \left | u + \sqrt{u ^{2} – 1} \right | + (C + C_{1}) \\ = \sqrt{u ^{2} – 1} – log \left | u + \sqrt{u ^{2} – 1} \right | + C_{2}$

Question 8:

Obtain an integral (or anti – derivative) of the $\frac{u ^{2}}{\sqrt{u ^{6} + m ^{6}}}$

Suppose, u 3 = z

3 u2 du = dz

$\int \frac{u ^{2}}{\sqrt{u ^{6} + m ^{6}}}\; du = \frac{1}{3} \int \frac{dz}{\sqrt{z ^{2} + (m^{3}) ^{2}}} \\ = \frac{1}{3} log \left | z + \sqrt{z ^{2} + m^{6}} \right | + C \\ = \frac{1}{3} log \left | u ^{3} + \sqrt{u ^{6} + m^{6}} \right | + C$

Question 9:

Obtain an integral (or anti – derivative) of the $\frac{sec ^{2}\; u}{\sqrt{tan ^{2} \;u + 4}}$

Suppose, tan u = z

sec 2 u du  = dz

$\int \frac{sec ^{2}\; u}{\sqrt{tan ^{2} \;u + 4}}\; du = \int \frac{dz}{\sqrt{z ^{2} + 2 ^{2}}} \\ = log \left | z + \sqrt{z ^{2} + 4} \right | + C \\ = log \left | tan\; u + \sqrt{tan ^{2} \;u + 4} \right | + C$

Question 10:

Obtain an integral (or anti – derivative) of the $\frac{1}{\sqrt{u ^{2} + 2u + 2}}$

$\int \frac{1}{\sqrt{u ^{2} + u + 2}} du = \int \frac{1}{(u + 1) ^{2} + (1) ^{2}} \\ Suppose,\; u + 1 = z \\ du = dz => \int \frac{1}{\sqrt{u ^{2} + 2 u + 2}} du = \int \frac{1}{\sqrt{z ^{2} + 1}} \;dz \\ = log \left | z + \sqrt{z ^{2} + 1} \right | + C \\ = log \left | (u + 1) + \sqrt{(u + 1) ^{2} + 1} \right | + C \\ = log \left | (u + 1) + \sqrt{u ^{2} + 2 u + 1} \right | + C$

Question 11:

Obtain an integral (or anti – derivative) of the $\frac{1}{\sqrt{9 u ^{2} + 6 u + 2}}$

$\int \frac{1}{\sqrt{9 u ^{2} + 6 u + 2}} du = \int \frac{1}{(3 u + 1) ^{2} + (2) ^{2}} \\ Suppose,\; 3 u + 1 = z \\ 3\; du = dz => \int \frac{1}{\sqrt{9 u ^{2} + 6 u + 2}} du = \frac{1}{3} \int \frac{1}{\sqrt{t ^{2} + 2 ^{2}}} \;dz \\ = \frac{1}{3} \left [ \frac{1}{2} tan ^{- 1} (\frac{z}{2}) \right ] + C \\ = \frac{1}{3} \left [ \frac{1}{2} tan ^{- 1} (\frac{3 u + 1}{2}) \right ] + C$

Question 12:

Obtain an integral (or anti – derivative) of the $\frac{1}{\sqrt{7 – 6 u – u ^{2}}}$

7 – 6 u – u2 = can also be written as 7 – (u2 + 6 u + 9 – 9)

Therefore,

7 – (u2 + 6 u + 9 – 9)

= 16 – (u2 + 6 u + 9)

= 16 – (u + 3) 2

= 4 2 – (u + 3) 2

$\int \frac{1}{\sqrt{7 – 6 u – u ^{2}}} du = \int \frac{1}{4 ^{2} – (u + 3) ^{2}} du \\ Suppose,\; u + 3 = z \\ du = dz \\ \int \frac{1}{4 ^{2} – (u + 3) ^{2}} du = \int \frac{1}{4 ^{2} – (z) ^{2}} dz \\ = sin ^{- 1} (\frac{z}{4}) + C \\ = sin ^{- 1} (\frac{u + 3}{4}) + C$

Question 13:

Obtain an integral (or anti – derivative) of the $\frac{1}{\sqrt{(u – 1) (u – 2)}}$

(u – 1) (u – 2) can be written as u 2 – 3 u + 2

Therefore,

u 2 – 3 u + 2

$= u ^{2} – 3 u + \frac{9}{4} – \frac{9}{4} + 2 \\ = \left ( u – \frac{3}{2} \right ) ^{2} – \frac{1}{4} \\ = \left ( u – \frac{3}{2} \right ) ^{2} – \left ( \frac{1}{2} \right ) ^{2} \\$ $\int \frac{1}{\sqrt{(u – 1) (u – 2)}}\; du = \int \frac{1}{\sqrt{\left ( u – \frac{3}{2} \right ) ^{2} – \left ( \frac{1}{2} \right ) ^{2}}} \;du \\ Suppose,\; u – \frac{3}{2} = z \\ du = dz \\$ $\int \frac{1}{\sqrt{\left ( u – \frac{3}{2} \right ) ^{2} – \left ( \frac{1}{2} \right ) ^{2}}} \;du = \int \frac{1}{\sqrt{z ^{2} – (\frac{1}{2}) ^{2}}} dz \\ = log \left | z + \sqrt{z ^{2} – (\frac{1}{2}) ^{2}} \right | + C \\ = log \left | (u – \frac{3}{2}) + \sqrt{(u – \frac{3}{2}) ^{2} – (\frac{1}{2}) ^{2}} \right | + C \\ = log \left | (u – \frac{3}{2}) + \sqrt{u^{2} – 3 u + 2} \right | + C$

Question 14:

Obtain an integral (or anti – derivative) of the $\frac{1}{\sqrt{8 + 3 u – u^{2}}}$

$\frac{1}{\sqrt{8 + 3 u – u^{2}}} can\; also\; be\; written\; as\; 8 – \left ( u^{2} – 3 u + \frac{9}{4} – \frac{9}{4} \right ) \\ Therefore,\; \frac{41}{4} – \left ( u – \frac{3}{2} \right ) ^{2} \\ \int \frac{1}{\sqrt{8 + 3 u – u^{2}}}\; du = \int \frac{1}{\sqrt{\frac{41}{4} – \left ( u – \frac{3}{2} \right ) ^{2}}}\; du \\$ $Suppose\; u – \frac{3}{2} = z \\ du = dz \\ \int \frac{1}{\sqrt{\frac{41}{4} – \left ( u – \frac{3}{2} \right ) ^{2}}}\; du = \frac{1}{\sqrt{\left ( \frac{\sqrt{41}}{2} \right ) ^{2} – z ^{2} }} \; dz \\ = sin ^{- 1} \left ( \frac{z}{\frac{\sqrt{41}}{2}} \right ) + C \\ = sin ^{- 1} \left ( \frac{u – \frac{3}{2}}{\frac{\sqrt{41}}{2}} \right ) + C \\ = sin ^{- 1} \left ( \frac{2 u – 3}{\sqrt{41}} \right ) + C \\$

Question 15:

Obtain an integral (or anti – derivative) of the $\frac{1}{\sqrt{(u – m) (u – n)}}$

$(u – m) (u – n)\; can\; also\; be\; written\; as\; u ^{2} – (m + n) u + mn \\ Therefore,\; \\ u ^{2} – (m + n) u + mn \\ = u ^{2} – (m + n) u + \frac{(m + n) ^{2}}{4} – \frac{(m + n) ^{2}}{4} + mn \\ = \left [ u – \frac{(m + n)}{2} \right ] ^{2} – \frac{(m – n) ^{2}}{4} \\ \int \frac{1}{\sqrt{(u – m) (u – n)}}\; du = \int \frac{1}{\sqrt{\left \{u – \frac{(m + n)}{2} \right \} ^{2} – \left ( \frac{(m – n)}{2} \right ) ^{2}}} \;du \\$ $Suppose,\; u – \left ( \frac{(m + n)}{2} \right ) = z \\ du = dz \\ \int \frac{1}{\sqrt{\left \{u – \frac{(m + n)}{2} \right \} ^{2} – \left ( \frac{(m – n)}{2} \right ) ^{2}}} \;du = \int \frac{1}{\sqrt{z ^{2} – \left ( \frac{m – n}{2} \right ) ^{2}}} \;dz \\$ $log\; \left |z + \sqrt{z ^{2} – (\frac{m – n}{2}) ^{2}} \right | + C \\ log\; \left | \left \{ u – (\frac{m + n}{2}) \right \} + \sqrt{(u – m) (u – n)} \right | + C$

Question 16:

Obtain an integral (or anti – derivative) of the $\frac{4 u + 1}{\sqrt{2 u^{2} + u – 3}}$

$Suppose,\; 4u + 1 = A \frac{d}{dx} (2 u^{2} + u – 3) + B …. (1) \\ 4 u + 1 = A (4u + 1) + B \\ 4 u + 1 = 4 Au + A + B$

Equate the coefficients of u and the constants on both the sides, we get

4 A = 4 => A = 1

A + B = 1 => B = 0

From (1), we get

Suppose, 2 u2 + u – 3 = z

(4 u + 1) du = dz

$\int \frac{4 u + 1}{\sqrt{2 u^{2} + u – 3}} du = \int \frac{1}{\sqrt{z}} \;dz \\ = 2 \sqrt{z} + C \\ = 2 \sqrt{2 u^{2} + u – 3} + C$

Question 17:

Obtain an integral (or anti – derivative) of the $\frac{u + 2}{\sqrt{u ^{2} – 1}}$

$Suppose,\; u + 2 = A \frac{d}{du} (u ^{2} – 1) + B …. (1) \\ u + 2 = A (2 u) + B$

Equate the coefficients of u and the constants on both the sides, we get

2 A = 1 => A = (1 / 2)

B = 2

From (1), we get

$From\; (1), we\; get, \\ (u + 2) = \frac{1}{2} (2 u) + 2 \\ Then,\; \int \frac{u + 2}{\sqrt{u ^{2} – 1}}\; du = \int \frac{\frac{1}{2} (2 u) + 2}{\sqrt{u ^{2} – 1}} \\ = \frac{1}{2} \int \frac{2 u}{\sqrt{u ^{2} – 1}} \;du + \int \frac{2}{\sqrt{u ^{2} – 1}} \;du ….. (2) \\$ $In\; \frac{1}{2} \int \frac{2 u}{\sqrt{u ^{2} – 1}} \;du = \frac{1}{2} \int \frac{dz}{\sqrt{z}} \\ = \frac{1}{2} [2 \sqrt{z}] + C \\ = \sqrt{z} + C \\ = \sqrt{u ^{2} – 1} \\ Then,\; \int \frac{2}{\sqrt{u ^{2} – 1}} \;du = 2 \int \frac{1}{\sqrt{u ^{2} – 1}} \;du \\ In\; equation\; (2), we\; get, \\ \int \frac{u + 2}{\sqrt{u ^{2} – 1}} \;du = \sqrt{u ^{2} – 1} + 2\; log\; \left | u + \sqrt{u ^{2} – 1} \right | + C$

Question 18:

Obtain an integral (or anti – derivative) of the $\frac{5u – 2}{1 + 2 u + 3 u^{2}}$

Suppose, $5 u – 2 = A \frac{d}{du} (1 + 2 u + 3 u^{2}) + B \\ 5 u – 2 = A (2 + 6 u) + B \\$

Equate the coefficients of u and the constants on both the sides, we get

$5 = 6 A => A = \frac{5}{6} \\ 2 A + B = – 2 => B = – \frac{11}{3} \\ 5 u – 2 = \frac{5}{6} (2 + 6 u) + \frac{- 11}{3} \\ \int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \int \frac{\frac{5}{6} (2 + 6 u) – \frac{11}{3}}{1 + 2 u + 3 u^{2}} \;du \\ = \frac{5}{6} \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du – \frac{11}{3} \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\$ $Suppose,\; I_{1} = \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du \; and\; I_{2} = \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\ \int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \frac{5}{6} I_{1} – \frac{11}{3} I_{2} …. (1) \\ I_{1} = \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du \;and\; I_{2} = \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\ Suppose,\; 1 + 2 u + 3 u^{2} = z \\ (2 + 6u)\; du = dz \\ I_{1} = \int \frac{dz}{z} \\ I_{1} = log \left | z \right | + C \\ I_{1} = log \left | 1 + 2 u + 3 u^{2} \right | + C ….. (2) I_{2} = \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\$ $1 + 2 u + 3 u^{2} can\; also\; be\; written\; as\; 1 + 3 \left ( u ^{2} + \frac{2}{3}\; u \right ) \\ Therefore, \; \\ 1 + 3 \left ( u ^{2} + \frac{2}{3}\; u \right ) \\ = 1 + 3 \left ( u ^{2} + \frac{2}{3}\; u + \frac{1}{9} – \frac{1}{9} \right ) \\ = 1 + 3 \left ( u + \frac{1}{3} \right ) ^{2} – \frac{1}{3} \\ = \frac{2}{3} + 3 \left ( u + \frac{1}{3} \right ) ^{2} \\ = 3 [\left ( u + \frac{1}{3} \right ) ^{2} + \frac{2}{9}] \\ = 3 [\left ( u + \frac{1}{3} \right ) ^{2} + (\frac{\sqrt{2}}{3}) ^{2}] \\$ $I_{2} = \frac{1}{3} \int \left [ \frac{1}{[\left ( u + \frac{1}{3} \right ) ^{2} + (\frac{\sqrt{2}}{3}) ^{2}]} \right ] du \\ = \frac{1}{3} \left [ \frac{1}{\frac{\sqrt{2}}{3}}\; tan ^{- 1} (\frac{u + \frac{1}{3}}{\frac{\sqrt{2}}{3}}) \right ] \\ = \frac{1}{3} \left [ \frac{3}{\sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) \right ] + C \\ = \frac{1}{\sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) + C ….. (3) \\$

Substituting equations (2) and (3) in equation (1), we get,

$\int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \frac{5}{6} \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du – \frac{11}{3} \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\ \int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \frac{5}{6} \left [ log \left | 1 + 2 u + 3 u^{2} \right | \right ] – \frac{11}{3} \left [\frac{1}{\sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) \right ] + C \\ = \frac{5}{6} log \left | 1 + 2 u + 3 u^{2} \right ] – \frac{11}{3 \sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) + C$

Question 19:

Obtain an integral (or anti – derivative) of the $\frac{6 u + 7}{\sqrt{(u – 5) (u – 4)}}$

Suppose, $\frac{6 u + 7}{\sqrt{(u – 5) (u – 4)}} = \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \\ Suppose,\; 6 u + 7 = A \frac{d}{du} (u ^{2} – 9 u + 20) + B \\ 6 u + 7 = A (2 u – 9) + B$

Equate the coefficients of u and the constants on both the sides, we get,

2 A = 6 => A = 3

–  9 + B = 7 => B = 34

6 u + 7 = 3 (2 u – 9) + 34

$\int \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \;du = \int \frac{3 (2 u – 9) + 34}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ = 3 \int \frac{2 u – 9}{\sqrt{u ^{2} – 9 u + 20}} \;du + 34 \int \frac{1}{\sqrt{u ^{2} – 9 u + 20}} \;du$ $Suppose,\; I_{1} = \int \frac{2 u – 9}{\sqrt{u ^{2} – 9 u + 20}} \;du \;and \;I_{2} = \int \frac{1}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ \int \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \;du = 3 I_{1} + 34 I_{2} \\ I_{1} = \int \frac{2 u – 9}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ Suppose,\; u ^{2} – 9 u + 20 = z (2 u – 9) \;du = dz \\ I_{1} = \frac{dz}{\sqrt{z}} \\ I_{1} = 2 \sqrt{z} \\ I_{1} = 2 \sqrt{u ^{2} – 9 u + 20} \\ and\; I_{2} = \int \frac{1}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ u ^{2} – 9 u + 20 \;can\; also \;be\; written\; as\; u ^{2} – 9 u + 20 + \frac{81}{4} – \frac{81}{4}$ $Therefore,\; u ^{2} – 9 u + 20 + \frac{81}{4} – \frac{81}{4} \\ = \left ( u – \frac{9}{2} \right ) ^{2} – \frac{1}{4} \\ = \left ( u – \frac{9}{2} \right ) ^{2} – (\frac{1}{2}) ^{2} \\ I_{2} = \int \frac{1}{\sqrt{\left ( u – \frac{9}{2} \right ) ^{2} – (\frac{1}{2}) ^{2}}} \;du \\ I_{2} = log \left |(u – \frac{9}{2}) + \sqrt{u ^{2} – 9 u + 20} \right | + C ….. (3)$

Substituting equations (2) and (3) in equation (1), we get,

$\int \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \;du = 3 [2 \sqrt{u ^{2} – 9 u + 20}] + 34\; log \left [ \left ( (u – \frac{9}{2}) + \sqrt{u ^{2} – 9 u + 20} \right ) \right ] + C \\ = 6 \sqrt{u ^{2} – 9 u + 20} + 34\; log \left [ \left ((u – \frac{9}{2}) + \sqrt{u ^{2} – 9 u + 20} \right ) \right ] + C$

Question 20:

Obtain an integral (or anti – derivative) of the $\frac{u + 2}{\sqrt{4u – u ^{2}}}$

Suppose, $u + 2 = A \frac{d}{du} (4u – u ^{2}) + B \\ (u + 2) = A (4 – 2 u) + B$

Equate the coefficients of u and the constants on both the sides, we get,

$- 2 A = 1 => A = – \frac{1}{2} \\ 4 A + B = 2 => B = 4 \\ (u + 2) = – \frac{1}{2} (4 – 2u) + 4 \\ \int \frac{u + 2}{\sqrt{4u – u ^{2}}} \;du = \int \frac{- \frac{1}{2}\; (4 – 2u) + 4}{\sqrt{4 u – u^{2}}} \;du \\ = – \frac{1}{2} \int \frac{4 – 2 u}{\sqrt{4 u – u^{2}}} \;du + 4 \int \frac{1}{\sqrt{4 u – u^{2}}} \;du$ $Suppose,\; I_{1} = \int \frac{4 – 2 u}{\sqrt{4 u – u^{2}}} \;du\; and\; I_{2} = \int \frac{1}{\sqrt{4 u – u^{2}}} \;du \\ \int \frac{u + 2}{\sqrt{4u – u ^{2}}} \;du = – \frac{1}{2} I_{1} + 4 I_{2} ….. (1) \\ Then,\; I_{1} = \int \frac{4 – 2 u}{\sqrt{4 u – u^{2}}} \;du \\ Suppose,\; 4 u – u ^{2} = z \\ (4 – 2 u) \;du = dz \\ I_{1} = \frac{dz}{\sqrt{z}} = 2 \sqrt{z} = 2\; \sqrt{4u – u ^{2}} ….. (2) \\ I_{2} = \int \frac{1}{\sqrt{4 u – u^{2}}} \;du \\ Suppose,\; 4 u – u^{2} = – (- 4 u + u^{2}) \\ (4 – 2 u) \;du = – (- 4 u + u^{2} + 4 – 4) \\ = 4 – (u – 2) ^{2} \\ I_{2} = \int \frac{1}{\sqrt{(2) ^{2} – (u – 2) ^{2}}} \;du = sin ^{- 1} (\frac{u – 2}{2}) ….. (3) \\$

Substituting equations (2) and (3) in equation (1), we get,

$\int \frac{u + 2}{\sqrt{4u – u ^{2}}} \;du = – \frac{1}{2} (2 \sqrt{4 u – u ^{2}}) + 4 sin^{- 1} \left ( \frac{u – 2}{2} \right ) + C \\ = – \sqrt{4u – u ^{2}} + 4 sin^{- 1} \left ( \frac{u – 2}{2} \right ) + C$

Question 21:

Obtain an integral (or anti – derivative) of the $\frac{u + 2}{\sqrt{4u ^{2} + 2 u + 3}}$

$\int \frac{u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du = \frac{1}{2} \int \frac{2 (u + 2)}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ = \frac{1}{2} \int \frac{2 u + 4}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ = \frac{1}{2} \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du + \frac{1}{2} \int \frac{2}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ = \frac{1}{2} \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du + \int \frac{1}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ Suppose,\; I_{1} = \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du \;and\; I_{2} = \int \frac{1}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ \int \frac{u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du = \frac{1}{2} I_{1} + I_{2} \\$ $I_{1} = \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ Suppose,\; u ^{2} + 2 u + 3 = z \\ (2 u + 2) \;du = dz \\ I_{1} = \int \frac{dz}{\sqrt{z}} = 2 \sqrt{z} = 2 \sqrt{u ^{2} + 2 u + 3} ….. (2) \\ I_{2} = \int \frac{1}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ u ^{2} + 2 u + 3 = u ^{2} + 2 u + 1 + 2 = (u + 1) ^{2} + (\sqrt{2}) ^{2} \\ I_{2} = \int \frac{1}{\sqrt{(u + 1) ^{2} + (\sqrt{2}) ^{2}}} \;du = log \left | (u + 1) + \sqrt{u ^{2} + 2 u + 3} \right | ….. (3)$

Substituting equations (2) and (3) in equation (1), we get,

$\int \frac{u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du = \frac{1}{2} (2 \sqrt{u ^{2} + 2 u + 3}) + log\; \left | (u + 1) + \sqrt{u ^{2} + 2 u + 3} \right | + C \\ = \sqrt{u ^{2} + 2 u + 3} + log \left | (u + 1) + \sqrt{u ^{2} + 2 u + 3} \right | + C$

Question 22:

Obtain an integral (or anti – derivative) of the $\frac{u + 2}{\sqrt{u ^{2} – 2 u – 5}}$

$Suppose,\; (u + 3) = A \frac{d}{du} (u ^{2} – 2u – 5) \\ (u + 3) = A (2 u – 2) + B \\$

Equate the coefficients of u and the constants on both the sides, we get,

$2 A = 1 => A = \frac{1}{2} \\ – 2 A + B = 3 => B = 4 \\ (u + 3) = \frac{1}{2} (2 u – 2) + 4 \\ \int \frac{u + 3}{\sqrt{u ^{2} – 2 u – 5}} \;du = \int \frac{\frac{1}{2} (2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du \\ = \frac{1}{2} \int \frac{(2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du + 4 \int \frac{1}{u ^{2} – 2 u – 5} du \\$ $Suppose,\; I_{1} = \int \frac{(2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du \;and\; I_{2} = \int \frac{1}{u ^{2} – 2 u – 5} du \\ \int \frac{u + 3}{\sqrt{u ^{2} – 2 u – 5}} \;du = \frac{1}{2} I_{1} + 4 I_{2} …. (1) \\ Then, I_{1} = \int \frac{(2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du \\ Suppose,\; u ^{2} – 2 u – 5 = Z \\ (2 u – 2) \;du = dz \\ I_{1} = \int \frac{dz}{z} \\ = log \left | z \right | + C \\ = log \left | u ^{2} – 2 u – 5 \right | + C …. (2)$ $I_{2} = \int \frac{1}{u ^{2} – 2 u – 5} du \\ = \int \frac{1}{(u ^{2} – 2 u + 1) – 6} du \\ = \int \frac{1}{(u – 1) ^{2} + (\sqrt{6}) ^{2}} \;du \\ = \frac{1}{2 \sqrt{6}} log \left ( \frac{u – 1 – \sqrt{6}}{u – 1 + \sqrt{6}} \right ) …… (3)$

Using equations (2) and (3) in equation (1), we get,

$\int \frac{u + 3}{\sqrt{u ^{2} – 2 u – 5}} \;du = \frac{1}{2} log \left | u ^{2} – 2 u – 5 \right | + 4 \left [ \frac{1}{2 \sqrt{6}} log \left ( \frac{u – 1 – \sqrt{6}}{u – 1 + \sqrt{6}} \right ) \right ] + C \\ = \frac{1}{2} log \left | u ^{2} – 2 u – 5 \right | + \frac{2}{\sqrt{6}} log \left ( \frac{u – 1 – \sqrt{6}}{u – 1 + \sqrt{6}} \right )$

Question 23:

Obtain an integral (or anti – derivative) of the $\frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}}$

$Suppose,\; 5u + 3 = A \frac{d}{du} (u ^{2} + 4 u + 10) + B \\ 5u + 3 = A (2 u + 4) + B$

Equate the coefficients of u and the constants on both the sides, we get,

$2 A = 5 => A = \frac{5}{2} \\ 4 A + B = 3 => B = – 7 \\ 5 u + 3 = \frac{5}{2} (2 u + 4) – 7 \\ \int \frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}} \;du = \frac{\frac{5}{2} ((2 u + 4) – 7)}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ = \frac{5}{2} \int \frac{2 u + 4}{\sqrt{u ^{2} + 4 u + 10}} \;du – 7 \int \frac{1}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ Suppose,\; I_{1} = \int \frac{2 u + 4}{\sqrt{u ^{2} + 4 u + 10}} \;du \;and\; I_{2} = \int \frac{1}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ \int \frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}} \;du = \frac{5}{2} I_{1} – 7 I_{2} \;\; ….. (1)$ $I_{1} = \int \frac{2 u + 4}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ Suppose,\; u ^{2} + 4 u + 10 = z \\ (2 u + 4) \;du = dz \\ I_{1} = \int \frac{dz}{\sqrt{z}} = 2 \sqrt{z} = 2 \sqrt{u ^{2} + 4 u + 10} ….. (2) I_{2} = \int \frac{1}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ = \int \frac{1}{\sqrt{(u ^{2} + 4 u + 4) + 6}} \;du \\ = \int \frac{1}{(u + 2) ^{2} + (\sqrt{6}) ^{2}} \;du \\ =log \left |(u + 2)\sqrt{u ^{2} + 4 u + 10} \right | ….. (3)$

Substituting equations (2) and (3) in equation (1), we get,

$\int \frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}} \;du = \frac{5}{2} [2 \sqrt{u ^{2} + 4 u + 10}] – 7 log \left | (u + 2) + \sqrt{u ^{2} + 4 u + 10} \right | + C \\ = 5\; \sqrt{u ^{2} + 4 u + 10} – 7 log \left | (u + 2) + \sqrt{u ^{2} + 4 u + 10} \right | + C \\$

Question 24: Which of the following below is the answer for $\int \frac{du}{u ^{2} + 2 u + 2} \;du$ $(a) u tan ^{- 1} (u + 1) + C \\ (b) tan ^{- 1} (u + 1) + C \\ (c) (u + 1) tan ^{- 1} (u) + C \\ (d) tan ^{- 1} (u) + C$

$\int \frac{du}{u ^{2} + 2 u + 2} \;du = \int \frac{du}{(u ^{2} + 2 u + 1) + 1} \\ = \int \frac{1}{(u + 1) ^{2} + (1) ^{2}} \;du \\ = \left [ tan ^{- 1} (u + 1) \right ] + C$

Thus, (b) is the correct answer.

Question 25: Which of the following below is the answer for $\int \frac{du}{\sqrt{9 u – 4 u ^{2}}} \;du$ $(a) \frac{1}{9} sin ^{- 1} \frac{9 u – 8}{8} + C \\ (b) \frac{1}{2} sin ^{- 1} \frac{8 u – 9}{9} + C \\ (c) \frac{1}{3} sin ^{- 1} \frac{9 u – 8}{8} + C \\ (d) \frac{1}{2} sin ^{- 1} \frac{9 u – 8}{8} + C$

$\int \frac{du}{\sqrt{9 u – 4 u ^{2}}} \;du \\ = \int \frac{du}{\sqrt{- 4 \left ( u ^{2} – \frac{9}{4} u \right )}} \\ = \int \frac{du}{\sqrt{- 4 \left ( u ^{2} – \frac{9}{4} u + \frac{81}{64} – \frac{81}{64} \right )}} \\ = \int \frac{1}{\sqrt{- 4} \left [ \left ( u – \frac{9}{8} \right ) ^{2} – (\frac{9}{8}) ^{2} \right ]} \;du \\ = \frac{1}{2} \int \frac{1}{\sqrt{(\frac{9}{8}) ^{2} – \left ( u – \frac{9}{8} \right ) ^{2}}} \;du \\ = \frac{1}{2} \left [ sin ^{- 1} \left ( \frac{u – \frac{9}{8}}{\frac{9}{8}} \right ) \right ] + C \\ = \frac{1}{2} sin ^{- 1} \left ( \frac{8 u – 9}{9} \right ) + C$

Thus, (b) is the correct answer.

Exercise 7.4

Question 1:

Obtain an integral (or anti – derivative) of the following rational number $\frac{u}{(u + 1) (u + 2)}$

Suppose, $\frac{u}{(u + 1) (u + 2)} = \frac{A}{u + 1} + \frac{B}{u + 2} \\ => u = A (u + 2) + B (u + 1)$

Equate the coefficients of u and the constants on both the sides, we get,

A + B = 1

2 A + B = 0

On solving, we get,

A = – 1 and B = 2

$\frac{u}{(u + 1) (u + 2)} = \frac{- 1}{u + 1} + \frac{2}{u + 2} \\ => \int \frac{u}{(u + 1) (u + 2)} \;du = \frac{- 1}{u + 1} + \frac{2}{u + 2} \;du \\ = – log \left | u + 1 \right | + 2 log \left | u + 2 \right | + C \\ = log \left ( u + 2 \right ) ^{2} – log \left | u + 1 \right | + C \\ = log \frac{(u + 1) ^{2}}{(u + 1)} + C$

Question 2:

Obtain an integral (or anti – derivative) of the following rational number $\frac{1}{u ^{2} – 9}$

Suppose, $\frac{1}{(u + 3) (u – 3)} = \frac{A}{u + 3} + \frac{B}{u – 3} \\ 1 = A (u – 3) + B (u + 3)$

Equate the coefficients of u and the constants on both the sides, we get,

A + B = 0

– 3 A + 3 B = 1

On solving, we get

$A = – \frac{1}{6} \; and\; B = \frac{1}{6} \\ \frac{1}{(u + 3) (u – 3)} = \frac{- 1}{6 (u + 3)} + \frac{1}{6 (u – 3)} \\ => \int \frac{1}{u ^{2} – 9} \;du = \int \left ( \frac{- 1}{6 (u + 3)} + \frac{1}{6 (u – 3)} \right ) \;du \\ = – \frac{1}{6} log \left | u + 3 \right | + \frac{1}{6} log \left | u – 3 \right | + C \\ = \frac{1}{6} log \left | \frac{(u – 3)}{(u + 3)} \right | + C$

Question 3:

Obtain an integral (or anti – derivative) of the following rational number $\frac{3 u – 1}{(u – 1) (u – 2) (u – 3)}$

Suppose, $\frac{3 u – 1}{(u – 1) (u – 2) (u – 3)} = \frac{A}{(u – 1)} + \frac{B}{(u – 2} + \frac{C}{(u – 3)} \\ 3 u – 1 = A (u – 2) (u – 3) + B (u – 1) (u – 3) + C (u – 1) (u – 2) ….. (1)$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– 5 A – 4 B – 3 C = 1

6 A + 3 B + 2 C = – 1

On solving, we get,

A = 1, B = – 5, and C = 4

$\frac{3 u – 1}{(u – 1) (u – 2) (u – 3)} = \frac{1}{(u – 1)} – \frac{5}{(u – 2} + \frac{4}{(u – 3)} \\ \int \frac{3 u – 1}{(u – 1) (u – 2) (u – 3)} \;du = \int \left \{ \frac{1}{(u – 1)} – \frac{5}{(u – 2} + \frac{4}{(u – 3)} \right \} \;du \\ = log \left | u – 1 \right | – 5 log \left | u – 2 \right | + 4 log \left | u – 3 \right | + C$

Question 4:

Obtain an integral (or anti – derivative) of the following rational number $\frac{u}{(u – 1) (u – 2) (u – 3)}$

Suppose, $\frac{u}{(u – 1) (u – 2) (u – 3)} = \frac{A}{(u – 1)} + \frac{B}{(u – 2} + \frac{C}{(u – 3)} \\ u = A (u – 2) (u – 3) + B (u – 1) (u – 3) + C (u – 1) (u – 2) ….. (1)$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– 5 A – 4 B – 3 C = 1

6 A + 3 B + 2 C = 0

On solving, we get,

$A = \frac{1}{2}, \;B = – 2 \;and \;C = \frac{3}{2} \\ \frac{u}{(u – 1) (u – 2) (u – 3)} = \frac{1}{2 (u – 1)} – \frac{2}{(u – 2} + \frac{3}{2 (u – 3)} \\ \int \frac{u}{(u – 1) (u – 2) (u – 3)} \;du = \int \left \{ \frac{1}{2 \;(u – 1)} – \frac{2}{(u – 2)} + \frac{3}{2 (u – 3)} \right \} \;du \\ = \frac{1}{2}\; log \left | u – 1 \right | – 2 log \left | u – 2 \right | + \frac{3}{2} log \left | u – 3 \right | + C$

Question 5:

Obtain an integral (or anti – derivative) of the following rational number $\frac{2 u}{u ^{2} + 3 u + 2}$

Suppose, $\frac{2 u}{u ^{2} + 3 u + 2} = \frac{A}{(u + 1)} + \frac{B}{(u + 2)} \\ 2 u = A (u + 2) + B (u + 1) …. (1)$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B = 2

2 A + B = 0

On solving, we get,

A = – 2 and B = 4

$\frac{2 u}{(u + 1) (u + 2)} = \frac{- 2}{(u + 1)} + \frac{4}{(u + 1)} \\ \int \frac{2 u}{(u + 1) (u + 2)} \;du = \int \left \{ \frac{4}{(u + 1)} – \frac{2}{(u + 1)}\right \} \;du \\ = 4 log \left | u + 2 \right | – 2 log \left | u + 1 \right | + C$

Question 6:

Obtain an integral (or anti – derivative) of the following rational number $\frac{1 – u ^{2}}{u (1 – 2 u)}$

$\frac{1 – u ^{2}}{u (1 – 2 u)}$ is not a proper fraction.

Dividing (1 – u 2) by u (1 – 2 u), we get,

$\frac{1 – u ^{2}}{u (1 – 2 u)} = \frac{1}{2} + \frac{1}{2} \left ( \frac{2 – u}{u (1 – 2 u)} \right ) \\ Suppose,\; \frac{2 – u}{u (1 – 2 u)} = \frac{A}{u} + \frac{B}{(1 -2 u)} \\ (2 – u) = A (1 – 2 u) + B u ……. (1)$

Equate the coefficients of u2, u and the constants on both the sides, we get,

– 2 A + B = – 1

And A = 2

On solving, we get,

A = 2 and B = 3

$\frac{2 – u}{u (1 – 2 u)} = \frac{2}{u} + \frac{3}{(1 -2 u)} \\ Using\; in\; equation\; (1), we\; get, \\ \frac{1 – u ^{2}}{u (1 – 2 u)} = \frac{1}{2} + \frac{1}{2} \left \{ \frac{2}{u} + \frac{3}{(1 -2 u)} \right \} \\ = \frac{u}{2} + log \left | u \right | + \frac{3}{2 (- 2)}\; log \left | 1 – 2u \right | + C \\ = \frac{u}{2} + log \left | u \right | – \frac{3}{4}\; log \left | 1 – 2u \right | + C$

Question 7:

Obtain an integral (or anti – derivative) of the following rational number $\frac{u}{(u ^{2} + 1) (u – 1)}$

Suppose, $\frac{u}{(u ^{2} + 1) (u – 1)} = \frac{A u + B}{(u ^{2} + 1)} + \frac{C}{u – 1} \\ u = (A u + B) (u – 1) + C (u^{2} + 1) \\ u = A u ^{2} – A u + B u – B + C u^{2} + C$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A  + C = 0

– A + B = 1

– B + C = 0

On solving, we get,

$A = – \frac{1}{2},\; B = \frac{1}{2}, and\; C = \frac{1}{2} \\ Using\; equation\; (1),\; we\; get\; \\ \frac{u}{(u ^{2} + 1) (u – 1)} = \frac{\left ( – \frac{1}{2} u + \frac{1}{2} \right )}{(u ^{2} + 1)} + \frac{\frac{1}{2}}{(u – 1)} \\$ $\int \frac{u}{(u ^{2} + 1) (u – 1)} \;du = – \frac{1}{2} \int \frac{u}{(u ^{2} + 1)} \;du + \frac{1}{2} \int \frac{1}{(u ^{2} + 1)} + \frac{1}{2} \int \frac{1}{2} \int \frac{1}{u – 1} \;du \\ = – \frac{1}{4} \int \frac{2 u}{(u ^{2} + 1)} + \frac{1}{2} tan ^{- 1} u + \frac{1}{2} log \left | u – 1 \right | + C \\ \int \frac{2 u}{(u ^{2} + 1)} \;du, let\; (u ^{2} + 1) = z => 2u\; du = dz \\ \int \frac{2 u}{(u ^{2} + 1)} \;du = \int \frac{dz}{z} = log \left | z \right | = log\; \left | u^{2} + 1 \right | \\ \int \frac{u}{(u ^{2} + 1) (u – 1)} \;du = – \frac{1}{4} log \left | (u ^{2} + 1) \right | + \frac{1}{2} tan ^{- 1}u + \frac{1}{2} log \left | u – 1 \right | + C \\ = \frac{1}{2} log \left | u – 1 \right | – \frac{1}{4} log \left | (u ^{2} + 1) \right | + \frac{1}{2} tan ^{- 1}u + C$

Question 8:

Obtain an integral (or anti – derivative) of the following rational number $\frac{u}{(u – 1)^{2} (u + 2)}$

$\frac{u}{(u – 1) c^{2} (u + 2)} = \frac{A}{(u – 1)} + \frac{B}{(u – 1) ^{2}} + \frac{C}{u + 2} \\ u = A (u – 1) (u + 2) + B (u + 2) + C (u – 1)^{2} \\$

Putting u = 1, we get,

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + C = 0

A + B – 2 C = 1

– 2 A + 2 B + C = 0

On solving, we get,

$A = \frac{2}{9},\; B = \frac{1}{3} \;and\; C = – \frac{2}{9}$ $\frac{u}{(u – 1)^{2} (u + 2)} = \frac{2}{9 \;(u – 1)} + \frac{1}{3 \;(u – 1) ^{2}} – \frac{2}{9 (u + 2)} \\ \int \frac{u}{(u – 1)^{2} (u + 2)} \;du = \frac{2}{9} \int \frac{1}{(u – 1)} \;du + \frac{1}{3} \int \frac{1}{(u – 1) ^{2}} \;du – \frac{2}{9} \int \frac{1}{(u + 2)} \;du \\ = \frac{2}{9} log\; \left | u – 1 \right | + \frac{1}{3} \left ( \frac{- 1}{u – 1} \right ) – \frac{2}{9} log \left | u + 2 \right | + C \\ = \frac{2}{9} log\; \left | \frac{u – 1}{u + 2} \right | – \frac{1}{3 (u – 1)} + C$

Question 9:

Obtain an integral (or anti – derivative) of the following rational number $\frac{3 u + 5}{u^{3} – u^{2} – u + 1}$

$\frac{3 u + 5}{u^{3} – u^{2} – u + 1} = \frac{3 u + 5}{(u – 1)^{2} (u + 1)} \\ Suppose,\; \frac{3 u + 5}{(u – 1)^{2} (u + 1)} = \frac{A}{(u – 1)} + \frac{B}{(u – 1)^{2}} + \frac{C}{(u + 1)} \\ 3 u + 5 = A (u – 1) (u + 1) + B (u + 1) + C (u – 1) ^{2} \\ 3 u + 5 = A (u^{2} – 1) + B (u + 1) + C (u^{2} + 1 – 2 u) \\$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + C = 0

B – 2 C = 3

– A + B + C = 5

On solving, we get,

B = 4

$A = – \frac{1}{2} \;and\; C = \frac{1}{2} \\ \frac{3 u + 5}{(u – 1)^{2} (u + 1)} = \frac{- 1}{2 (u – 1)} + \frac{4}{(u – 1)^{2}} + \frac{1}{2 (u + 1)} \\ \int \frac{3 u + 5}{(u – 1)^{2} (u + 1)} \;du = – \frac{1}{2} \int \frac{1}{(u – 1)} \;du + 4 \int \frac{1}{(u – 1)^{2}} \;du + \frac{1}{2} \int \frac{1}{(u + 1)} \;du \\ = – \frac{1}{2} log\; \left | u – 1 \right | + 4 \left ( \frac{- 1}{u – 1} \right ) + \frac{1}{2} log \left | u + 1 \right | + C \\ = \frac{1}{2} log \left | \frac{u + 1}{u – 1} \right | – \frac{4}{(u – 1)} + C$

Question 10:

Obtain an integral (or anti – derivative) of the following rational number $\frac{2 u – 3}{(u^{2} – 1) (2u + 3)}$

$\frac{2 u – 3}{(u^{2} – 1) (2u + 3)} = \frac{2 u – 3}{(u + 1)(u – 1)(2u + 3)}$

Suppose, $\frac{2 u – 3}{(u^{2} – 1) (2u + 3)} = \frac{A}{(u + 1)} + \frac{B}{(u – 1)} + \frac{C}{(2u + 3)} \\ (2u – 3) = A\; (u – 1)(2 u + 3) + B\; (u + 1) (2 u + 3) + C\; (u + 1) (u – 1) \\ (2u – 3) = A\; (2 u^{2} + u – 3) + B\; (2 u^{2} + 5u + 3) + C\; (u^{2} – 1) \\ (2u – 3) = (2A + 2B + C) u^{2} + (A + 5B)u + (- 3A + 3B – C)$

Equate the coefficients of u2, u and the constants on both the sides, we get,

2A + 2B + C = 1

A + 5B = 2

– 3A + 3B – C = – 3

On solving, we get,

$\frac{2 u – 3}{(u + 1)(u – 1)(2u + 3)} = \frac{5}{2 (u + 1)} – \frac{1}{10 (u – 1)} – \frac{24}{5 (2u + 3)} \\ \frac{2 u – 3}{(u + 1)(u – 1)(2u + 3)} = \frac{5}{2} \int \frac{1}{(u + 1)} \;du – \frac{1}{10} \int \frac{1}{ (u – 1)} \;du – \frac{24}{5} \int \frac{1}{ (2u + 3)} \;du \\ = \frac{5}{2} log\; \left | u + 1 \right | – \frac{1}{10} log\; \left | u – 1 \right | – \frac{24}{5 \times 2} log\; \left | 2u + 3 \right | + C \\ = \frac{5}{2} log\; \left | u + 1 \right | – \frac{1}{10} log\; \left | u – 1 \right | – \frac{12}{5} log\; \left | 2u + 3 \right | + C$

Question 11:

Obtain an integral (or anti – derivative) of the following rational number $\frac{5 u}{(u + 1) (u^{2} – 4)}$

$\frac{5 u}{(u + 1) (u^{2} – 4)} = \frac{5 u}{(u + 1)(u + 2)(u – 2)} \\ Suppose,\; \frac{5 u}{(u + 1)(u + 2)(u – 2)} = \frac{A}{(u + 1)} + \frac{B}{(u + 2)} + \frac{C}{(u – 2)} \\ 5u = A (u + 2)(u – 2) + B (u + 1)(u – 2) + C (u + 1)(u + 2) …. (1)$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– B + 3C = 5 and

–  4A – 2B + 2C = 0

On solving, we get,

$A = \frac{5}{3}, B = – \frac{5}{2} \;and\; C = \frac{5}{6} \\ \frac{5 u}{(u + 1)(u + 2)(u – 2)} = \frac{5}{3 (u + 1)} – \frac{5}{2 (u + 2)} + \frac{5}{6 (u – 2)} \\ \int \frac{5 u}{(u + 1)(u + 2)(u – 2)} \;du = \frac{5}{3} \frac{1}{(u + 1)} \;du – \frac{5}{2} \frac{1}{(u + 2)} \;du + \frac{5}{6} \frac{1}{(u – 2)} \;du \\ = \frac{5}{3} log\; \left | u + 1 \right | – \frac{5}{2} log\; \left | u + 2 \right | + \frac{5}{6} log\; \left | u – 2 \right | + C$

Question 12:

Obtain an integral (or anti – derivative) of the following rational number $\frac{u^{3} + u + 1}{u^{2} – 1}$

$\frac{u^{3} + u + 1}{u^{2} – 1}$ is not a proper fraction.

So, dividing (u3 + u + 1) by u2 – 1, we get,

$\frac{u^{2} + u + 1}{u^{2} – 1} = u + \frac{2u + 1}{u^{2} – 2} \\ Suppose,\; \frac{2u + 1}{u^{2} – 2} = \frac{A}{(u + 1)} + \frac{B}{(u – 1)} \\ 2u + 1 = A (u – 1) + B (u + 1) ….. (1)$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B = 2

– A + B = 1

On solving, we get,

$A = \frac{1}{2} \;and\; B = \frac{3}{2} \\ \frac{u^{2} + u + 1}{u^{2} – 1} = u + \frac{1}{2 (u + 1)} + \frac{3}{2 (u – 1)} \\ Integrating\; on\; both\; the\; sides,\; we\; get, \int \frac{u^{2} + u + 1}{u^{2} – 1} \;du = \int u \;du + \frac{1}{2} \int \frac{1}{u + 1} \;du + \frac{3}{2} \frac{1}{(u – 1)} \;du \\ = \frac{u ^{2}}{2} + \frac{1}{2} log\; \left | u + 1 \right | – \frac{3}{2} log\; \left | u – 1 \right | + C$

Question 13:

Obtain an integral (or anti – derivative) of the following rational number $\frac{2}{(1 – u)(1 + u ^{2})}$

Suppose, $\frac{2}{(1 – u)(1 + u ^{2})} = \frac{A}{1 – u} + \frac{Bu + C}{1 + u ^{2}} \\ 2 = A (1 + u ^{2}) + (Bu + C)(1 – u) \\ 2 = A + A\; u^{2} + Bu – B\; u^{2} + C – Cu$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A – B = 0

B – C = 0

A + C = 2

On solving, we get,

A = 1, B = 1 and C = 1

$\frac{2}{(1 – u)(1 + u ^{2})} = \frac{1}{1 – u} + \frac{u + 1}{1 + u ^{2}} \\ \int \frac{2}{(1 – u)(1 + u ^{2})} \;du = \int \frac{1}{1 – u} \;du + \int \int \frac{u}{1 + u ^{2}} \;du + \int \frac{1}{1 + u ^{2}} \;du \\ = – log\; \left | u – 1 \right | + \frac{1}{2} log\; \left | 1 + u^{2} \right | + tan ^{- 1} \;u + C$

Question 14:

Obtain an integral (or anti – derivative) of the following rational number $\frac{3u – 1}{(u + 2)^{2}}$

Suppose, $\frac{3u – 1}{(u + 2)^{2}} = \frac{A}{(u + 2)} + \frac{B}{(u + 2)^{2}} \\ 3u – 1 = A (u + 2) + B$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = 3

2A + B = – 1

B = – 7

$\frac{3u – 1}{(u + 2)^{2}} = \frac{3}{(u + 2)} – \frac{7}{(u + 2)^{2}} \\ 3u – 1 = A (u + 2) + B \frac{3u – 1}{(u + 2)^{2}} = 3 \int \frac{1}{(u + 2)} \;du – 7 \int \frac{u}{(u + 2)^{2}} \;du \\ = 3\; log\; \left | u + 2 \right | – 7 \left ( \frac{- 1}{(u\; + \;2)} \right ) + C \\ = 3\; log\; \left | u + 2 \right | + \left ( \frac{7}{(u\; + \;2)} \right ) + C$

Question 15:

Obtain an integral (or anti – derivative) of the following rational number $\frac{1}{u^{4} – 1}$

$\frac{1}{u^{4} – 1} = \frac{1}{(u^{2} – 1)(u^{4} + 1)} = \frac{1}{(u + 1) (u – 1)(1 + u^{2})} \\ Suppose,\; \frac{1}{(u + 1) (u – 1)(1 + u^{2})} = \frac{A}{(u + 1)} + \frac{B}{(u – 1)} + \frac{Cu + D}{(u^{2} + 1)} \\ 1 = A (u + 1)(u^{2} + 1) + B (u + 1)(u^{2} + 1) + (Cu + D) (u^{2} – 1) \\ 1 = A (u^{3} + u – u^{2} – 1) + B (u^{3} + u + u^{2} + 1) + Cu^{2} + Du^{2} – Cu – D \\ 1 = (A + B + C) u^{3} + (- A + B + D) u^{2} + (A + B – C)u + (- A + B – D)$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– A + B + D = 0

A + B – C = 0

– A + B – D = 1

On solving, we get,

$A = – \frac{1}{4}, B = \frac{1}{4}, C = 0 \;and\; D = – \frac{1}{2} \\ \frac{1}{u^{4} – 1} = \frac{- 1}{4 (u + 1)} + \frac{1}{4 (u – 1)} – \frac{1}{2 (u^{2} + 1)} \\ \int \frac{1}{u^{4} – 1} \;du = – \frac{1}{4} \int \frac{1}{(u + 1)} \;du + \frac{1}{4} \int \frac{1}{(u – 1)} \;du + \frac{1}{2} \int \frac{1}{(u^{2} + 1)} \\ = – \frac{1}{4} log\; \left | u + 1 \right | + \frac{1}{4} log\; \left | u – 1 \right | – \frac{1}{2} tan^{- 1} \;u + C \\ = \frac{1}{4} log\; \left | \frac{ u – 1}{u + 1} \right | – \frac{1}{2} tan^{- 1} \;u + C$

Question 16:

Obtain an integral (or anti – derivative) of the following rational number $\frac{1}{u (u^{m} + 1)}$ [Hint: multiply denominator and numerator by u n – 1 and put un = z]

$\frac{1}{u (u^{m} + 1)}$

Multiplying denominator and numerator by u n – 1, we get,

$\frac{1}{u (u^{m} + 1)} = \frac{u ^{m – 1}}{u ^{m – 1}. u (u^{m} + 1)} = \frac{u ^{m – 1}}{u ^{m} (u^{m} + 1)} \\ Suppose,\; u^{m} = z => u ^{m – 1} \;du = dz \\ \int \frac{1}{u (u^{m} + 1)} \;du = \int \frac{u ^{m – 1}}{u ^{m} (u^{m} + 1)} \;du = \frac{1}{m} \int \frac{1}{z (z + z)} \;du \\ Suppose,\; \frac{1}{z (z + z)} = \frac{A}{z} + \frac{B}{(z + 1)} \\ 1 = A (1 + z) + Bz \\$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = 1 and B = – 1

$\frac{1}{z (z + z)} = \frac{1}{z} – \frac{1}{(z + 1)} \\ \int \frac{1}{u (u^{m} + 1)} \;du = \frac{1}{m} \int \left \{ \frac{1}{z} – \frac{1}{(z + 1)} \right \} + C \\ = \frac{1}{m} \left [ log\; \left | u^{m} \right | – log\; \left | u^{m} + 1 \right | \right ] + C \\ = \frac{1}{m} log\; \left | \frac{u^{m}}{u^{m} + 1} \right |$

Question 17:

Obtain an integral (or anti – derivative) of the following rational number $\frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)}$ [Hint: Put sin u = z]

$\frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)}$

Suppose, sin u = z => cos u du = dz

$\int \frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)} \;du = \int \frac{dz}{(1 – z)(2 – z)} \\ Suppose,\; \frac{1}{(1 – z)(2 – z)} = \frac{A}{(1 – z)} + \frac{B}{(2 – z)} \\ 1 = A (2 – z) + B (1 – z)$

Equate the coefficients of u2, u and the constants on both the sides, we get,

– 2A – B = 0

2A + B = 1

On solving, we get,

A = 1 and B = – 1

$\frac{1}{(1 – z)(2 – z)} = \frac{1}{(1 – z)} – \frac{1}{(2 – z)} \\ \int \frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)} \;du = \int \left \{ \frac{1}{(1 – z)} – \frac{1}{(2 – z)} \right \} \;dz \\ = – log\; \left | 1 – z \right | + log\; \left | 2 – z \right | + C \\ = log\; \left | \frac{2 – z}{1 – z} \right | + C \\ = log\; \left | \frac{2 – sin\; u}{1 – sin\; u} \right | + C$

Question 18:

Obtain an integral (or anti – derivative) of the following rational number $\frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)}$

$\frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)} = 1 – \frac{(4 u^{2} + 10)}{(u^{2} + 3)(u^{2} + 4)} \\ Suppose,\; \frac{(4 u^{2} + 10)}{(u^{2} + 3)(u^{2} + 4)} = \frac{Au + B}{(u^{2} + 3)} + \frac{Cu + D}{(u^{2} + 4)} \\ (4 u^{2} + 10) = (Au + B)(u^{2} + 4) + (Cu + D)(u^{2} + 3) \\ (4 u^{2} + 10) = A u^{3} + 4 Au + B u^{2} + 4 B + C u^{3} + 3 Cu + D u^{2} + 3D \\ (4 u^{2} + 10) = (A + C) u^{3} + (B + D) u^{2} + (4A + 3C) u + (4B + 3D)$

Equate the coefficients of u3, u2, u and the constants on both the sides, we get,

A + C = 0

B + D = 4

4 A + 3 C = 0

4 B + 3 D = 10

On solving, we get,

A = 0, B = – 2, C = 0 and D = 6

$\frac{(4 u^{2} + 10)}{(u^{2} + 3)(u^{2} + 4)} = \frac{- 2}{(u^{2} + 3)} + \frac{6}{(u^{2} + 4)} \\ \frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)} = 1 – \left ( \frac{- 2}{(u^{2} + 3)} + \frac{6}{(u^{2} + 4)} \right ) \\ \int \frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)} \;du = \int \left \{ 1 + \frac{2}{(u^{2} + 3)} + \frac{6}{(u^{2} + 4)} \right \} \\ = \int \left \{ 1 + \frac{2}{u ^{2} + (\sqrt{3})^{2}} – \frac{6}{u^{2} + 2^{2}} \right \} \;du \\ = u + 2 \left ( \frac{1}{\sqrt{3}} \;tan ^{- 1} \frac{u}{\sqrt{3}} \right ) – 6 \left ( \frac{1}{2} \;tan ^{- 1} \frac{u}{2} \right ) + C \\ = u + \frac{2}{\sqrt{3}} \;tan ^{- 1} \frac{u}{\sqrt{3}} – 3 \;tan ^{- 1} \frac{u}{2} + C$

Question 19:

Obtain an integral (or anti – derivative) of the following rational number $\frac{2\; u}{(u^{2} + 1)(u^{2} + 3)}$

$\frac{2\; u}{(u^{2} + 1)(u^{2} + 3)}$

Suppose, u2 = z

2u du = dz

$\int \frac{2\; u}{(u^{2} + 1)(u^{2} + 3)} \;du = \int \frac{dz}{(z + 1)(z + 3)} …. (1) \\ Suppose,\; \frac{1}{(z + 1)(z + 3)} = \frac{A}{(z + 1)} + \frac{B}{(z + 3)} \\ 1 = A\; (z + 3) + B\; (z + 1) …. (1) \\$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A  + B = 0

3 A + B = 1

On solving, we get,

$A = \frac{1}{2} \;and\; B = – \frac{1}{2} \\ \frac{1}{(z + 1)(z + 3)} = \frac{1}{2\; (z + 1)} – \frac{1}{2\; (z + 3)} \\ \int \frac{2\; u}{(u^{2} + 1)(u^{2} + 3)} \;du = \int \left \{ \frac{1}{2\; (z + 1)} – \frac{1}{2\; (z + 3)} \right \} \;dz \\ = \frac{1}{2} log \left | (z + 1) \right | – \frac{1}{2} log \left | (z + 3) \right | + C \\ = \frac{1}{2} log \left | \frac{z + 1}{z + 3} \right | + C \\ = \frac{1}{2} log \left | \frac{u^{2} + 1}{u^{2} + 3} \right | + C$

Question 20:

Obtain an integral (or anti – derivative) of the following rational number $\frac{1}{u (u^{4} – 1)}$

$\frac{1}{u (u^{4} – 1)}$

Multiplying denominator and numerator by u3, we get,

$\frac{1}{u (u^{4} – 1)} = \frac{u^{3}}{u^{4} (u^{4} – 1)} \\ \int \frac{1}{u (u^{4} – 1)} \;du = \int \frac{u^{3}}{u^{4} (u^{4} – 1)} \;du \\ Suppose,\; u^{4} = t => 4 u^{3} \;du = dz \\ \int \frac{1}{u (u^{4} – 1)} \;du = \frac{1}{4} \int \frac{dz}{z (z – 1)} \\ Suppose,\; \frac{1}{z (z – 1)} = \frac{A}{z} + \frac{B}{z – 1} \\ 1 = A (z – 1) + Bz ….. (1) \\$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = – 1 and B = 1

$\frac{1}{z (z – 1)} = \frac{- 1}{z} + \frac{1}{z – 1} \\ = \int \frac{1}{u (u^{4} – 1)} \;du = \frac{1}{4} \int \left ( \frac{- 1}{z} + \frac{1}{z – 1} \right ) \;dz \\ = \frac{1}{4} \left [ – log\; \left | z \right | + log\; \left | z – 1 \right | \right ] + C \\ = \frac{1}{4} log\; \left | \frac{z – 1}{z} \right | + C \\ = \frac{1}{4} log\; \left | \frac{u^{4} – 1}{u^{4}} \right | + C$

Question 21:

Obtain an integral (or anti – derivative) of the following rational number $\frac{1}{(e^{u} – 1)}$

$\frac{1}{(e^{u} – 1)}$

Suppose, eu = z

eu du = dz

$\int \frac{1}{(e^{u} – 1)} \;du = \int \frac{1}{z – 1} \times \frac{dz}{z} = \int \frac{1}{z (z – 1)} \;dz \\ Suppose,\; \frac{1}{z (z – 1)} = \frac{A}{z} + \frac{B}{z – 1} \\ 1 = A (z – 1) + Bz$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = – 1 and B = 1

$\frac{1}{z (z – 1)} = \frac{- 1}{z} + \frac{1}{z – 1} \\ \int \frac{1}{z (z – 1)} \;du = log\; \left | \frac{z – 1}{z} \right | + C \\ = log\; \left | \frac{e^{u} – 1}{e^{u}} \right | + C$

Question 22: Which of the following below is an integral of $\frac{u \;du}{(u – 1)(u – 2)}$ $(a) log\; \left | \frac{(u – 1)^{2}}{u – 2} \right | + C \\ (b) log\; \left | \frac{(u – 2)^{2}}{u – 2} \right | + C \\ (c) log\; \left | (\frac{(u – 1)}{u – 2}) ^{2} \right | + C \\ (d) log\; \left | (u – 1)(u – 2) \right | + C$

$Suppose,\; \frac{u \;du}{(u – 1)(u – 2)} = \frac{A}{(u – 1)} + \frac{B}{u – 2} \\ u = A (u – 2) + B (u – 1) …. (1)$

Equate the coefficients of u and the constants on both the sides, we get,

A = – 1 and B = 2

$\frac{u \;du}{(u – 1)(u – 2)} = \frac{- 1}{(u – 1)} + \frac{2}{u – 2} \\ \int \frac{u \;du}{(u – 1)(u – 2)} \;d= \left \{ \frac{- 1}{(u – 1)} + \frac{2}{u – 2} \right \} \;du \\ = – log\; \left | u – 1 \right | + 2\; log\; \left | u – 2 \right | + C \\ = log\; \left | \frac{(u – 2) ^{2}}{u – 1} \right | + C$

Hence, option (b) is the correct answer.

Question 23: Which of the following below is an integral of $\int \frac{du}{u (u^{2} + 1)} \;du$ $(a) log\; \left | u \right | – \frac{1}{2} log\; (u^{2} + 1) + C \\ (b) log\; \left | u \right | + \frac{1}{2} log\; (u^{2} + 1) + C \\ (c) – log\; \left | u \right | + \frac{1}{2} log\; (u^{2} + 1) + C \\ (d) log\; \left | u \right | + \frac{1}{2} log\; (u^{2} + 1) + C \\$

$Suppose,\; \frac{1}{u (u^{2} + 1)} = \frac{A}{u} + \frac{Bu + C}{u^{2} + 1} \\ 1 = A (u^{2} + 1) + (Bu + C) u$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B = 0

C = 0

A = 1

On solving, we get,

A =1, B = – 1, and C = 0

$\frac{1}{u (u^{2} + 1)} = \frac{1}{u} + \frac{- U}{u^{2} + 1} \\ \int \frac{1}{u (u^{2} + 1)} \;du = \int \left \{ \frac{1}{u} – \frac{u}{u^{2} + 1} \right \} \;du \\ = log\; \left | u \right | – \frac{1}{2} log\; \left | u^{2} + 1 \right | + C$

Hence, option (a) is the correct answer.

Exercise 7.6

Question 1:

Obtain an integral of  u sin u.

Suppose, I = $\int u\; sin\; u \;du$

Integrating the equation by parts by taking u as first function and sin u as second function, we get,

$I = u\;\int sin\; u \;du – \int \left \{ \left ( \frac{d}{du} \;u \right ) \int sin\; u \;du \right \} \;du \\ = u (- cos\; u) – \int 1. (- cos\; u) \;du = – u\; cos\; u + sin\; u + C$

Question 2:

Obtain an integral of  u sin 3u.

Suppose, I = $\int u\; sin\; u\; du$

Integrating the equation by parts by taking u as first function and sin 3u as second function, we get,

$I = u\;\int sin\; 3u \;du – \int \left \{ \left ( \frac{d}{du} \;u \right ) \int sin\; 3u \;du \right \} \;du \\ = u (\frac{- cos\; 3u}{3}) – \int 1. (\frac{- cos\; 3u}{3}) \;du \\ = \frac{- u\; cos\; 3u}{3} + \frac{1}{9} sin\; 3u + C$

Question 3:

Obtain an integral of $u^{2}. e^{u}$

Suppose, I = $\int u^{2}. e^{u} \;du$

Integrating the equation by parts by taking u2 as first function and eu as second function, we get,

$I = u^{2} \int e^{u} \;du – \int \left \{ \left ( \frac{d}{du} u^{2} \right ) \int e^{u} \;du \right \} \;du \\ = u^{2} e^{u} – \int 2u\; e^{u} \;du \\ = u^{2} e^{u} – 2 \int u\; e^{u} \;du \\ Integrating\; by\; parts\;, we\; get, \\ = u^{2} e^{u} – 2 \left [u\; \int e^{u} \;du – \int \left \{ \left ( \frac{d}{du}\; u \right ). \int e^{u} \;du \right \} \right ] \;du \\ = u^{2} e^{u} – 2 \left [u\; e^{u} – \int e^{u} \;du \right ] \\ = u^{2} e^{u} – 2 \left [u\; e^{u} – e^{u} \right ] \\ = u^{2} e^{u} – 2 u\; e^{u} – 2 e^{u} + C \\ = e^{u} (u^{2} – 2u + 2) + C$

Question 4:

Obtain an integral of u log u.

Suppose, I = $\int u\; log\; u \;du$

Integrating the equation by parts by taking log u as first function and u as second function, we get,

$I = log\; u \int u\; du – \int \left \{ \left ( \frac{d}{du} log\; u \right ) \int u \;du \right \} \;du \\ = log\; u. \frac{u^{2}}{2} – \int \frac{1}{u}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} log\; u}{2} – \int \frac{u}{2} \;du \\ = \frac{u^{2} log\; u}{2} = \frac{u^{2}}{4} + C$

Question 5:

Obtain an integral of u log 2u.

Suppose, I = $\int u\; log\; 2u \;du$

Integrating the equation by parts by taking log 2 u as first function and u as second function, we get,

$I = log\; 2u \int u\; du – \int \left \{ \left ( \frac{d}{du} 2 log\; u \right ) \int u \;du \right \} \;du \\ = log\; 2u. \frac{u^{2}}{2} – \int \frac{2}{2 u}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} log\; 2u}{2} – \int \frac{u}{2} \;du \\ = \frac{u^{2} log\; 2u}{2} – \frac{u^{2}}{4} + C$

Question 6:

Obtain an integral of u2 log u.

Suppose, I = $\int u^{2}\; log\; u \;du$

Integrating the equation by parts by taking log u as first function and u 2 as second function, we get,

$I = log\; u \int u ^{2} \; du – \int \left \{ \left ( \frac{d}{du} log\; u \right ) \int u^{2} \;du \right \} \;du \\ = log\; u. \frac{u^{3}}{3} – \int \frac{1}{u}. \frac{u^{2}}{3} \;du \\ = \frac{u^{3} log\; u}{3} – \int \frac{u ^{2}}{3} \;du \\ = \frac{u^{3} log\; u}{2} – \frac{u^{3}}{9} + C$

Question 7:

Obtain an integral of u sin – 1 u.

Suppose, I = $\int u\; sin ^{- 1}\; u \;du$

Integrating the equation by parts by taking sin – 1 u as first function and u as second function, we get,

$I = sin ^{- 1}\; u \int u \;du – \int \left \{ \left ( \frac{d}{du} sin ^{- 1}\; u \right ) \int u \;du \right \} \;du \\ = sin ^{- 1}\; u \frac{u^{2}}{2} – \int \frac{1}{\sqrt{1 – u^{2}}}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \int \frac{- u^{2}}{\sqrt{1 – u^{2}}} \;du \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \int \left \{ \frac{1 – u^{2}}{\sqrt{1 – u^{2}}} – \frac{1}{\sqrt{1 – u^{2}}} \right \} \;du$ $= \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \int \left \{ \sqrt{1 – u^{2}} – \frac{1}{\sqrt{1 – u^{2}}} \right \} \;du \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \left \{ \int \sqrt{1 – u^{2}} \;du – \int \frac{1}{\sqrt{1 – u^{2}}} \;du \right \} \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \left \{ \frac{u}{2} \sqrt{1 – u^{2}} + \frac{1}{2} sin ^{- 1} \;u – sin ^{- 1} \;u \right \} + C \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{u}{4} \sqrt{1 – u^{2}} + \frac{1}{4} sin ^{- 1} \;u – \frac{1}{2} sin ^{- 1} \;u + C \\ = \frac{1}{4} (2u^{2} – 1) sin ^{- 1} \;u + \frac{u}{4} \sqrt{1 – u ^{2}} + C$

Question 8:

Obtain an integral of u tan – 1 u

Suppose, I = $\int u \;tan ^{- 1}\; u\; du$

Integrating the equation by parts by taking tan – 1 u as first function and u as second function, we get,

$I = tan ^{- 1}\; u \int u \;du – \int \left \{ \left ( \frac{d}{du} tan ^{- 1}\; u \right ) \int u \;du \right \} \;du \\ = tan ^{- 1}\; u \;\frac{u^{2}}{2} – \int \frac{1}{1 + u^{2}}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int \frac{ u^{2}}{1 + u^{2}} \;du \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int \left \{ \frac{u^{2} + 1}{1 + u^{2}} – \frac{1}{1 + u^{2}} \right \} \;du$ $= \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int \left \{ 1 – \frac{1}{1 + u^{2}} \right \} \;du \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int (u – tan ^{- 1} \;u) + C \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{u}{2} + \frac{1}{2}\; tan ^{- 1}\; u + C$

Question 9:

Obtain an integral of u cos – 1 u

Suppose, I = $\int u\; cos ^{- 1}\; u\; du$

Integrating the equation by parts by taking cos – 1 u as first function and u as second function, we get,

$I = cos ^{- 1}\; u \int u \;du – \int \left \{ \left ( \frac{d}{du} cos ^{- 1}\; u \right ) \int u \;du \right \} \;du \\ = cos ^{- 1}\; u \frac{u^{2}}{2} – \int \frac{- 1}{\sqrt{1 – u^{2}}}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} cos ^{- 1}\; u}{2} – \frac{1}{2} \int \frac{1 – u^{2} + 1}{\sqrt{1 – u^{2}}} \;du \\ = \frac{u^{2} cos ^{- 1}\; u}{2} + \frac{1}{2} \int \left \{ \sqrt{1 – u^{2}} + \frac{- 1}{\sqrt{1 – u^{2}}} \right \} \;du \\ = \frac{u^{2} cos ^{- 1}\; u}{2} – \frac{1}{2} I_{1} – \frac{1}{2}\; cos ^{- 1} \;u ….. (1)$ $where\; I_{1} = \int \sqrt{1 – u^{2}} \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{d}{du}\; \sqrt{1 – u^{2}} \int u\; du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{- 2u}{2 \sqrt{1 – u^{2}}}. u \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{- u^{2}}{\sqrt{1 – u^{2}}} \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{1 – u^{2} – 1}{\sqrt{1 – u^{2}}} \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \left \{ \int \sqrt{1 – u^{2}} \;du + \int \frac{- du}{\sqrt{1 – u^{2}}} \right \}$ $I_{1} = u \;\sqrt{1 – u^{2}} – \left \{ I_{1} + cos ^{- 1} \;u \right \} \\ 2 I_{1} = u \;\sqrt{1 – u^{2}} – cos ^{- 1} \;u \\ I_{1} = \frac{u}{2} \;\sqrt{1 – u^{2}} – \frac{1}{2} cos ^{- 1} \;u \\ Using\; in\; equation\; (1),\; we\; get, \\ I = \frac{u^{2} sin ^{- 1}\; u}{2} – \frac{1}{2} \left ( \frac{u}{2}\; \sqrt{1 – u^{2}} – \frac{1}{2}\; cos ^{- 1} \;u \right ) – \frac{1}{2}\; cos ^{- 1} \;u \\ = \frac{(2 u^{2} – 1)}{4}\; cos ^{- 1} \;u – \frac{u}{4}\; \sqrt{1 – u^{2}} + C$

Question 10:

Obtain an integral of (sin – 1 u) 2

Suppose, I = $\int (sin ^{- 1} \;u) ^{2} . 1\; du$

Integrating the equation by parts by taking (sin – 1 u) 2 as first function and 1 as second function, we get,

$sin ^{- 1} \;u \int 1 \;du – \int \left \{ \frac{d}{du} (sin ^{- 1} \;u)^{2}. \int 1. du \right \} \;du \\ = (sin ^{- 1} \;u)^{2}. \;u – \int \frac{2\; (sin ^{- 1} \;u)}{\sqrt{1 – u^{2}}}. \;u \;du \\ = u. (sin ^{- 1} \;u)^{2} + \int sin ^{- 1} \;u. \left ( \frac{- 2u}{\sqrt{1 – u^{2}}} \right ) \;du \\ = u. (sin ^{- 1} \;u)^{2} + \left [sin ^{- 1} \;u \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du – \int \left \{ \left ( \frac{d}{du} sin ^{- 1} \;u \right ) \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du\right \} \;du \right ] \\ = u. (sin ^{- 1} \;u)^{2} + \left [ sin ^{- 1} \;u. 2\sqrt{1 – u^{2}} – \int \frac{1}{\sqrt{1 – u^{2}}} . 2 \sqrt{1 – u^{2}} \;du \right ]$ $= u. (sin ^{- 1} \;u)^{2} + 2 \sqrt{1 – u^{2}}\; sin^{- 1} \;u – \int 2 \;du \\ = u. (sin ^{- 1} \;u)^{2} + 2 \sqrt{1 – u^{2}}\; sin^{- 1} \;u -2u + C$

Question 11:

Obtain an integral of $\frac{u\; cos^{- 1} \;u}{\sqrt{1 – u^{2}}}$

Suppose, I = $\int \frac{u\; cos^{- 1} \;u}{\sqrt{1 – u^{2}}} \;du \\ I = \frac{- 1}{2} \int \frac{- 2u}{\sqrt{1 – u^{2}}}.\; cos ^{- 1} u \;du$

Integrating the equation by parts by taking cos – 1 u as first function and $\frac{- 2u}{\sqrt{1 – u^{2}}}$ as second function, we get,

$I = \frac{- 1}{2} \left [cos^{- 1} \;u \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du – \int \left \{ \left ( \frac{d}{du}\; cos^{- 1} \;u \right ) \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du \right \} \;du \right ] \\ = \frac{- 1}{2} \left [ cos^{- 1} \;u. 2\; \sqrt{1 – u^{2}} – \int \frac{- 1}{\sqrt{1 – u^{2}}}. 2\; \sqrt{1 – u^{2}} \;du \right ] \\ = \frac{- 1}{2} \left [2\; \sqrt{1 – u^{2}}\; cos^{- 1} \;u + \int 2 \;du \right ] \\ = \frac{- 1}{2} \left [ 2\; \sqrt{1 – u^{2}}\; cos^{- 1} \;u + 2u \right ] + C \\ = – \left [\sqrt{1 – u^{2}} \;cos^{- 1} \;u + u \right ] + C$

Question 12:

Obtain an integral of u sec 2 u

Suppose, I = $\int u\; sec^{2} \;u \;du$

Integrating the equation by parts by taking u as first function and sec 2 u as second function, we get,

$u \int sec^{2} \;u \;du – \int \left \{ \left \{ \frac{d}{du}. u \right \} \int sec^{2} \;u \;du \right \} \;du \\ = u\; tan\; u – \int 1. \;tan \;u \;du \\ = u\; tan\; u – log\; \left | cos\; u \right | + C$

Question 13:

Obtain an integral of tan – 1 u

Suppose, I = $\int tan ^{- 1} \;u \;du$

Integrating the equation by parts by taking tan – 1 u as first function and 1 as second function, we get,

$tan ^{- 1} \;u \int 1 \;du – \int \left \{ \left ( \frac{d}{du} \;tan ^{- 1} \;u \right ) \int 1 \;du \right \} \;du \\ = tan ^{- 1} \;u . \;u – \int \frac{1}{1 + u^{2}}. \;u \;du \\ = tan ^{- 1} \;u . \;u – \frac{1}{2} \int \frac{2u}{1 + u^{2}}. \;du \\ = u\; tan ^{- 1} \;u – \frac{1}{2} \;log\; \left | 1 + u^{2} \right | + C \\ = u\; tan ^{- 1} \;u – \frac{1}{2} \;log\; (1 + u^{2}) + C$

Question 14:

Obtain an integral of u (log u) 2.

Suppose, I =

Integrating the equation by parts by taking (log u) 2 as first function and 1 as second function, we get,

$I = (log \;u) ^{2} \int u \;du – \int \left [ \left \{ (\frac{d}{du} \; log \;u) ^{2} \right \} \int u\; du \right ] \;du \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \left [ \int 2 log\; u. \frac{1}{u}. \frac{u^{2}}{2} \;du \right ] \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \int u\; log\; u \;du$

Integrating the equation again by parts, we get,

$I = \frac{u^{2}}{2} (log \;u) ^{2} \int u \;du – \left [log \;u \int u \;du – \left \{ (\frac{d}{du} \; log \;u) \int u \;du \right \} \;du \right ] \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \left [\frac{u^{2}}{2} – log\; u – \int \frac{1}{u}. \frac{u^{2}}{2} \;du \right ] \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \frac{u^{2}}{2} (log \;u) + \frac{1}{2} \int u \;du \\ = \frac{u^{2}}{2} (log \;u)^{2} – \frac{u^{2}}{2} (log \;u) + \frac{u^{2}}{4} + C$

Question 15:

Obtain an integral of (u 2 + 1) log u

Suppose, I = $\int (u ^{2} + 1) log\; u\; du = \int u ^{2}\; log\; u\; du + \int log\; u\; du \\ Suppose,\; I = I_{1} + I_{2} + …… (1) \\ Where,\; I_{1} = \int u ^{2}\; log\; u\; du \;and\; I_{2} = \int log\; u\; du \\ I_{1} = \int u ^{2}\; log\; u\; du$

Integrating the equation by parts by taking u as first function and u 2 as second function, we get,

$I_{1} = (log \;u) – \int u^{2} \;du – \int \left \{ \left ( \frac{d}{du} log \;u \right ) \int u ^{2} \;du \right \} \;du \\ = log \;u. \frac{u ^{3}}{3} – \int \frac{1}{u}. \frac{u ^{3}}{3} \;du \\ = \frac{u ^{3}}{3}\; log \;u – \frac{1}{3} (\int u^{2} \;du) \;du \\ = \frac{u ^{3}}{3}\; log \;u – \frac{u ^{3}}{9} + C_{1} …. (2) \\ I_{2} = \int log\; u\; du$

Integrating the equation by parts by taking u as first function and u 2 as second function, we get,

$I_{2} = (log \;u) – \int 1 \;du – \int \left \{ \left ( \frac{d}{du} log \;u \right ) \int 1 \;du \right \} \\ = log \;u. u – \int \frac{1}{u}. u \;du \\ = u\; log \;u – \int 1 \;du \\ = u\; log \;u – u + C_{2} ….. (3)$

Substituting equations (2) and (3) in equation (1), we get,

$I = \frac{u ^{3}}{3}\; log \;u – \frac{u ^{3}}{9} + C_{1} + u\; log \;u – u + C_{2} \\ = \frac{u ^{3}}{3}\; log \;u – \frac{u ^{3}}{9} + u\; log \;u – u + (C_{1} + C_{2}) \\ = \left (\frac{u ^{3}}{3} + u \right ) log \;u – \frac{u ^{3}}{9} – C$

Question 16:

Obtain an integral of e u (sin u + cos u)

Suppose, I = $\int e^{u} (sin\; u + cos\; u) \; du \\ Suppose,\; f (u) = sin\; u \\ f’ (u) = cos\; u \\ I = \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du \\ As\; we\; know, \\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e^{u} f (u) + C \\ I = e^{u} \;sin\; u + C$

Question 17:

Obtain an integral of $\frac{e^{u}}{(1 + u) ^{2}}$

Suppose, I = $\int \frac{u \;e^{u}}{(1 + u) ^{2}} \;du = \int e^{u} \left \{ \frac{u}{(1 + u) ^{2}} \right \} \;du \\ = \int e^{u} \left \{ \frac{1 + u – 1}{(1 + u) ^{2}} \right \} \;du \\ = \int e^{u} \left \{ \frac{1}{1 + u} – \frac{1}{(1 + u) ^{2}} \right \} \;du \\ Suppose,\; f (u) = \frac{1}{1 + u}, \; \; \; f’ (u) = \frac{- 1}{(1 + u) ^{2}} \\ \int \frac{u\; e^{u}}{(1 + u) ^{2}} \;du = \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du \\ As\; we\; know,\; \\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u} \;f (u) + C \\ \int \frac{u\; e^{u}}{(1 + u) ^{2}} \;du = \frac{e^{u}}{1 + u} + C$

Question 18:

Obtain an integral of $e^{u} \left ( \frac{1 + sin\; u}{1 + cos\; u} \right )$

$e^{u} \left ( \frac{1 + sin\; u}{1 + cos\; u} \right ) \\ = e^{u} \left ( \frac{sin ^{2} \;\frac{u}{2} + cos ^{2} \;\frac{u}{2} + 2 sin \;\frac{u}{2} \;cos\; \frac{u}{2}}{2\; cos ^{2} \;\frac{u}{2}} \right ) \\ = \frac{e^{u} \left ( sin \;\frac{u}{2} + cos \;\frac{u}{2} \right ) ^{2}}{2 cos ^{2} \;\frac{u}{2}} \\ = \frac{1}{2} e^{u} \left ( \frac{sin \;\frac{u}{2} + cos \;\frac{u}{2}}{cos \;\frac{u}{2}} \right ) ^{2} \\ = \frac{1}{2} e^{u} \left [ tan \;\frac{u}{2} + 1 \right ] ^{2} \\$ $= \frac{1}{2} e^{u} \left [1 + tan \;\frac{u}{2} \right ] ^{2} \\ = \frac{1}{2} e^{u} \left [ 1 + tan ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \\ = \frac{1}{2} e^{u} \left [ sec ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \\ \frac{e^{u} (1 + sin\; u) \;du}{(1 + cos\; u)} = \left [ \frac{1}{2} sec ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \; \; \; ….. (1) \\ Suppose,\; tan\; \frac{u}{2} = f (u) \; so \; f’ (u) = \frac{1}{2} sec ^{2} \;\frac{u}{2}$

As we know,

$\int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u}\; f (u) + C$

Considering equation (1), we get,

$\int \frac{e^{u} (1 + sin\; u) \;du}{(1 + cos\; u)} = e^{u} tan\; \frac{u}{2} + C$

Question 19:

Obtain an integral of $e ^{u} \left ( \frac{1}{u} – \frac{1}{u ^{2}} \right )$

Suppose, I = $\int e ^{u} \left ( \frac{1}{u} – \frac{1}{u ^{2}} \right ) \; du \\ Suppose,\; \frac{1}{u} = f (u) \; \; \; f’ (u) = \frac{- 1}{u^{2}} \\ As\; we\; know,\\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u}\; f (u) + C \\ I = \frac{e ^{u}}{u} + C$

Question 20:

Obtain an integral of $\frac{(u – 3) e^{u}}{(u – 1) ^{3}}$

$\int e^{u} \left \{ \frac{(u – 3)}{(u – 1) ^{3}} \right \} \;du = \int e^{u} \left \{ \frac{(u – 3)}{(u – 1 – 2) ^{3}} \right \} \;du \\ = \int e^{u} \left \{ \frac{1}{(u – 1) ^{2} – \frac{2}{(u – 1) ^{3}}} \right \} \;du \\ f (u) = \frac{1}{(u – 1) ^{2}} \; \; \; f’ (u) = \frac{- 2}{(u – 1) ^{3}} \\ As\; we\; know, \\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u}\; f (u) + C \\ \int e^{u} \left \{ \frac{(u – 3)}{(u – 1) ^{3}} \right \} \;du = \frac{e ^{u}}{(u – 1) ^{2}} + C$