Ncert Solutions For Class 12 Maths Ex 7.2

Ncert Solutions For Class 12 Maths Chapter 7 Ex 7.2

Question 1:

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

Answer:

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 \(\frac{(log\; u) ^{2}}{u}\)

Answer:

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

Answer:

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

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

Answer:

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

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 \(sin\; (mr + n) cos\; (mr + n)\)

Answer:

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 \(\sqrt{mr + n}\)

Answer:

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 \(u \sqrt{u + 2}\)

Answer:

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 \(u \sqrt{1 + 2 u ^{2}}\)

Answer:

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 \((4 u + 2) \sqrt{u ^{2} + u + 1}\)

Answer:

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 \(\frac{1}{u – \sqrt{u}}\)

Answer:

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

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

Answer:

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 \((u ^{3} – 1) ^{\frac{1}{3}} u ^{5}\)

Answer:

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 \(\frac{u ^{2}}{(2 + 3u ^{3}) ^{3}} \\\)

Answer:

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 \(\frac{1}{u (log u) ^{n}}, x > 0 \\\)

Answer:

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 \(\frac{u}{9 – 4 u ^{2}}\)

Answer:

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 \(e ^{2 m + 3}\)

Answer:

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 \(\frac{u}{e ^{u ^{2}}}\)

Answer:

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 \(\frac{e ^{tan ^{- 1} \Theta}}{1 + \Theta ^{2}}\)

Answer:

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 \(\frac{e ^{2u} – 1}{e ^{2u} + 1}\)

Answer:

\(\frac{e ^{2u} – 1}{e ^{2u} + 1}\)

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 \(\frac{e ^{2u} – e ^{- 2u}}{e ^{2u} + e ^{- 2u}}\)

Answer:

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 \(tan ^{2} (2 \Theta – 3)\)

Answer:

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

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 \(sec ^{2} (7 – 4 \theta)\)

Answer:

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 \(\frac{sin ^{- 1} \theta}{\sqrt{1 – \theta ^{2}}}\)

Answer:

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

Answer:

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

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

Answer:

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

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 \(\frac{cos \sqrt{\theta }}{\sqrt{\theta }}\)

Answer:

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 \(\sqrt{sin\; 2 \theta}\; cos\; 2 \theta\)

Answer:

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 \(\frac{cos\; \theta}{\sqrt{1 + sin\; \theta}}\)

Answer:

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 \(cot\; \theta\; log\; sin\; \theta\)

Answer:

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 \(\frac{sin\; \theta}{1 + cos\; \theta}\)

Answer:

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 \(\frac{sin\; \theta}{(1 + cos\; \theta) ^{2}}\)

Answer:

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 \(\frac{1}{1 + cot\; \theta}\)

Answer:

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 \(\frac{1}{1 – tan \theta}\)

Answer:

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 \(\frac{\sqrt{tan \theta}}{sin\; \theta\; cos\; \theta}\)

Answer:

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 \(\frac{(1 + log u) ^{2}}{u}\)

Answer:

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 \(\frac{(u + 1)(u + log u) ^{2}}{u}\)

Answer:

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

Question 37:

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

Answer:

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

 

 

Question 38:

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

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

Answer:

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

Question 39:

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

\((a) tan\; u + cot\; u + C \\ (b) tan\; u – cot\; u + C \\ (c) tan\; u \; cot\; u + C \\ (d) tan\; u – cot\; 2u + C \\\)

Answer:

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

Therefore, B is the correct answer

 

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