Integration - Properties, Problems, Irrational Algebraic Functions

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Integration is the process of finding the antiderivative. The integration of g′(x), with respect to dx, is given by

∫ g′(x) dx = g(x) + C, where C is the constant of integration.

The two types of integrals are

Definite integral: An integral with limits, namely upper and lower limits, without the constant of integration.
Indefinite integral: An integral without limits and with an arbitrary constant.

This article covers standard integrals, properties of integration, important formulas and examples of integration which helps students to have a deep knowledge of the topic.

Standard Integrals

Integrals of Rational and Irrational Functions

\begin{array}{l} \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C, n \neq 1 \\ \int \frac{1}{x} d x = \ln | x | + C \\ \int c d x = c \cdot x + C \\ \int x d x = \frac{x^{2}}{2} + C \\ \int x^{2} d x = \frac{x^{3}}{3} + C \\ \int \frac{1}{x^{2}} d x = - \frac{1}{x} + C \end{array}

\begin{array}{l} \int \sqrt{x} d x = \frac{2 \cdot x \cdot \sqrt{x}}{3} + C \\ \int \frac{1}{1 + x^{2}} d x = \arctan x + C \\ \int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + C \end{array}

Integrals of Trigonometric Functions

\begin{array}{l} \int \sin x d x = - \cos x + C \\ \int \cos x d x = \sin x + C \\ \int \tan x d x = \ln | \sec x | + C \\ \int \sec x d x = \ln | \tan x + \sec x | + C \end{array}

\begin{array}{l} \int \sin^{2} x d x = \frac{1}{2} (x - \sin x \cdot \cos x) + C \\ \int \cos^{2} x d x = \frac{1}{2} (x + \sin x \cdot \cos x) + C \\ \int \tan^{2} x d x = \tan x - x + C \\ \int \sec^{2} x d x = \tan x + C \end{array}

Integrals of Exponential and Logarithmic Functions

\begin{array}{l} \int \ln x d x = x \cdot \ln x - x + C \\ \int x^{n} \cdot \ln x d x = \frac{x^{n + 1}}{n + 1} \ln x - \frac{x^{n + 1}}{(n + 1)^{2}} + C \\ \int e^{x} d x = e^{x} + C \\ \int a^{x} d x = \frac{a^{x}}{\ln a} + C \end{array}

Properties of Integration

Property 1:

\begin{array}{l} \int_{a}^{a} f (x) d x = 0 \end{array}

Property 2:

\begin{array}{l} \int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x \end{array}

Property 3:

\begin{array}{l} \int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t \end{array}

Property 4:

\begin{array}{l} \int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{a}^{b} f (x) d x \end{array}

Property 5:

\begin{array}{l} (i) \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \end{array}

\begin{array}{l} (i i) \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x \end{array}

⇒ Also Read: Definite and Indefinite Integration

Useful Formulas

$\begin{array}{l} \int e^{a x} \sin b x = \frac{e^{a x}}{a^{2} + b^{2}} [a \sin b x - b \cos b x] \end{array}$
$\begin{array}{l} \int e^{a x} \cos b x = \frac{e^{a x}}{a^{2} + b^{2}} [a \cos b x + b \sin b x] \end{array}$

$\begin{array}{l} \int e^{x} (f (x) + f^{'} (x)) = e^{x} f (x) \end{array}$

Illustration:

\begin{array}{l} \int e^{x} (\sin x + \cos x) d x = e^{x} \sin x + c \end{array}

\begin{array}{l} \int e^{x} (l n x + \frac{1}{x}) d x = e^{x} l n x + c \end{array}

Integration of Trigonometric Functions

Type 1:

\begin{array}{l} I = \int \sin^{m} x \cos^{n} x d x \end{array}

1. If m is odd, put cos x = t

2. If n is odd, put sin x = t

3. If m, n rationales then put tan x = t

4. If both are even, then use the reduction method.

\begin{array}{l} Q \int \frac{\cos^{3} x}{\sin^{6} x} d x = \int \frac{1 - t^{2}}{t^{6}} d t \end{array}

Where t = sin x

\begin{array}{l} = \int t^{- 6} - t^{- 4} d t \end{array}

\begin{array}{l} = - \frac{1}{5 s i n^{5} x} + \frac{1}{3 \sin^{3} x} + c \end{array}

Type 2:

\begin{array}{l} \int \frac{d x}{a \cos x + b \sin x + c} \end{array}

Put t = tan (x/2)

Illustration

\begin{array}{l} \int \frac{d x}{2 + \sin x} \end{array}

\begin{array}{l} \Rightarrow t = \tan (\frac{x}{2}) \end{array}

\begin{array}{l} d x = \frac{2 d t}{1 + t^{2}} \end{array}

\begin{array}{l} = \int \frac{\frac{2 d t}{1 + t^{2}}}{2 + \frac{2 t}{1 + t^{2}}} \end{array}

\begin{array}{l} \Rightarrow \int \frac{d t}{t^{2} + t + 1} \end{array}

\begin{array}{l} = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 t + 1}{\sqrt{3}}) \end{array}

\begin{array}{l} = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 \tan \frac{x}{2} + 1}{\sqrt{3}}) + c \end{array}

Some Useful Substitutions for Irrational Functions

Form 1 :
$\begin{array}{l} \int l i n e a r \sqrt{Q u a d r a t i c} d x \end{array}$

\begin{array}{l} Substitute l i n e a r = m Q u d r a t i c^{'} + n \end{array}

Form 2:
$\begin{array}{l} \int \frac{d x}{l i n \sqrt{l i n_{1}}}, \int \frac{l i n}{\sqrt{l i n_{1}}} d x, \int \frac{\sqrt{l i n_{1}}}{l i n} d x \end{array}$

\begin{array}{l} Substitute l i n_{1} = t^{2} \end{array}

Form 3:
$\begin{array}{l} \int \frac{1}{l i n \sqrt{Q u a}} d x \end{array}$

Substitute lin = 1/t

Form 4:
$\begin{array}{l} \int \frac{d x}{(a x^{2} + b) \sqrt{(x^{2} + d)}} \end{array}$

Substitute x = 1/t and then u² for at² + b

Integration Formulas

$\begin{array}{l} \int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t \end{array}$
$\begin{array}{l} \int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x \end{array}$
$\begin{array}{l} \int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x \end{array}$
$\begin{array}{l} \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \end{array}$
$\begin{array}{l} \int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x \end{array}$
$\begin{array}{l} = 0 if \\ , f (2 a - x) = - f (x) \end{array}$
and
$\begin{array}{l} = 2 \int_{0}^{a} f (x) if f (2 a - x) = f (x) \end{array}$
$\begin{array}{l} \int_{- a}^{a} f (x) d x = {\begin{array}{c} 0 & i f f (x) is odd \\ 2 \int_{0}^{a} f (x) d x & i f f (x) is even \end{array} \end{array}$

Problems on Integration

Illustration:

\begin{array}{l} \int_{0}^{2} x^{2} [x] d x = \int_{0}^{1} x^{2} [x] d x + \int_{1}^{2} x^{2} [x] d x \end{array}

\begin{array}{l} = \int_{0}^{1} x^{2} .0 d x + \int_{1}^{2} x^{2} [1] d x \end{array}

\begin{array}{l} = 0 + {\frac{x^{3}}{3} |}_{1}^{2} \end{array}

\begin{array}{l} = \frac{8 - 1}{3} = \frac{7}{3} \end{array}

Illustration:

\begin{array}{l} \int_{π / 6}^{π / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x \end{array}

\begin{array}{l} I = \int \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x \(by x \to \frac{π}{2} - x) \end{array}

\begin{array}{l} 2 I = \int_{π / 6}^{π / 3} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} = \int_{π / 6}^{π / 3} 1 d x = \frac{π}{6} \end{array}

\begin{array}{l} I = \frac{π}{12} \end{array}

Illustration:

\begin{array}{l} I = \int \sin^{100} x \cos^{99} x \end{array}

Here, f(2π – x) = f(x)

Or

\begin{array}{l} I = 2 \int_{0}^{π} \sin^{100} x \cos^{99} x \end{array}

\begin{array}{l} = 2 \int_{0}^{π} \sin^{100} (π - x) \cos^{99} (π - x) \end{array}

I = -I

I = 0

Illustration:

\begin{array}{l} \int_{- 5}^{5} x^{3} = 0 as f (x) is odd \end{array}

Leibnitz’s rule

\begin{array}{l} \frac{d}{d x} \int_{u (x)}^{v (x)} f (t) d t = f (v (x)) \frac{d v (x)}{d x} - f (u (x)) u^{'} (x) \end{array}

Practice Problems

Problem 1.

\begin{array}{l} If \int_{x^{2}}^{x^{3}} \frac{1}{\log t} d t = y find \end{array}

\begin{array}{l} \frac{d y}{d x} = x (x - 1) {(\log x)}^{- 1} \end{array}

Problem 2.

\begin{array}{l} If \int_{\sin x}^{1} t^{2} f (t) d t = 1 - \sin x . \end{array}

where x ∈ (0, π/2), find f(1/√3).

Problem 3.

\begin{array}{l} If f (2) = 6, f^{'} (2) = \frac{1}{48} . Find lim_{x \to 2} \int_{6}^{f (x)} \frac{4 t^{3}}{x - 2} d t = 18. \end{array}

Problem 4.

\begin{array}{l} lim_{x \to \infty} \frac{\int_{0}^{x} e^{x^{2}} d x}{\int_{0}^{x} e^{2 x^{2}} d x} = 0 \end{array}

Integration by Parts

\begin{array}{l} \int u v d x = u \int v d x - \int u^{'} (\int v d x) d x \end{array}

Illustration:

Q.

\begin{array}{l} \int ℓ n x = \int ℓ n x .1 d x \end{array}

\begin{array}{l} = x ℓ n x - \int \frac{1}{x} . x d x \end{array}

\begin{array}{l} = x ℓ n x - x \end{array}

Q.

\begin{array}{l} \int x e^{x} d x = x \int e^{x} d x - \int (1) (\int e^{x} d x) d x \end{array}

\begin{array}{l} = x e^{x} - \int e^{x} d x \end{array}

\begin{array}{l} = x e^{x} - e^{x} \end{array}

Integration of Irrational Algebraic Functions

Type

\begin{array}{l} \int \frac{d x}{{(a x + b)}^{k} \sqrt{p x + q}} \end{array}

Q.

\begin{array}{l} \int \frac{x}{(x - 3) \sqrt{x + 1}} d x \end{array}

Put x + 1 = t², we get

\begin{array}{l} I = \int \frac{(t^{2} - 1) 2 t d t}{(t^{2} - 4) t} = 2 \int \frac{t^{2} - 1}{t^{2} - 4} d t \end{array}

\begin{array}{l} = 2 \int 1 + \frac{3}{t^{2} - 4} d t \end{array}

\begin{array}{l} = 2 t + \frac{3}{2} ℓ n | \frac{t - 2}{t + 2} | + c \end{array}

\begin{array}{l} = 2 \sqrt{x + 1} + \frac{3}{2} ℓ n | \frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2} | + c \end{array}

$\begin{array}{l} \Rightarrow \int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x \end{array}$

= 0 if f(2a – x) = -f(x)

\begin{array}{l} = 2 \int_{0}^{a} f (x) if f (2 a - x) = f (x) \end{array}

Optimisation of Area for Greatest and Least Values

Illustration:

If area by y = f(x) and y = x² + 2 between abscissa x = 2 and x = α is α³ – 4α² + 8. Find f(x).

Answer:

\begin{array}{l} α^{3} - 4 α^{2} + 8 = \int_{2}^{α} (x^{2} + 2 - f (x)) d x \end{array}

Differentiating with Labniz equation,

\begin{array}{l} 3 α^{2} - 8 α = α^{2} + 2 - f (α) \end{array}

\begin{array}{l} f (x) = - 2 x^{2} + 8 x + 2 \end{array}

Integrations Important JEE Main Questions

732

Definite Integration JEE Questions

80,755

Indefinite Integration JEE Questions

77,284

Solved Problems on Integration

Problem 1:

\begin{array}{l} \int_{}^{} \frac{d x}{\cos (x - a) \cos (x - b)} = \end{array}

Solution:

\begin{array}{l} \int_{}^{} \frac{d x}{\cos (x - a) \cos (x - b)} \\ = \frac{1}{\sin (a - b)} \int_{}^{} \frac{\sin {(x - b) - (x - a)}}{\cos (x - a) . \cos (x - b)} d x \\ = \frac{1}{\sin (a - b)} \int_{}^{} {\frac{\sin (x - b)}{\cos (x - b)} - \frac{\sin (x - a)}{\cos (x - a)}} d x \\ = cosec (a - b) \log \frac{\cos (x - a)}{\cos (x - b)} + c \end{array}

Problem 2:

\begin{array}{l} \int_{}^{} \frac{d x}{\sqrt{x + a} + \sqrt{x + b}} = \end{array}

Solution:

\begin{array}{l} \int_{}^{} \frac{d x}{\sqrt{x + a} + \sqrt{x + b}} = \int_{}^{} \frac{\sqrt{x + a} - \sqrt{x + b}}{(x + a) - (x + b)} d x \\ = \frac{1}{(a - b)} \int_{}^{} {(x + a)}^{1 / 2} d x - \frac{1}{(a - b)} \int_{}^{} {(x + b)}^{1 / 2} d x \\ = \frac{2}{3 (a - b)} [{(x + a)}^{3 / 2} - {(x + b)}^{3 / 2}] + c \end{array}

Problem 3:

\begin{array}{l} \int_{}^{} \frac{x^{3} - x - 2}{(1 - x^{2})} d x = \end{array}

Solution:

\begin{array}{l} \int_{}^{} \frac{x^{3} - x - 2}{(1 - x^{2})} d x \\ = \int_{}^{} \frac{- x (1 - x^{2})}{(1 - x^{2})} d x - \int_{}^{} \frac{2}{1 - x^{2}} d x \\ = - \int_{}^{} x d x - 2 \int_{}^{} \frac{1}{1 - x^{2}} d x = \frac{- x^{2}}{2} + \log (\frac{x - 1}{x + 1}) + c . \end{array}

Problem 4:

\begin{array}{l} \int_{}^{} \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} d x = \end{array}

Solution:

\begin{array}{l} \int_{}^{} \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} d x \\ = \int_{}^{} \frac{(\sin^{4} x + \cos^{4} x) (\sin^{4} x - \cos^{4} x)}{{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x} d x \\ = \int_{}^{} (\sin^{4} x - \cos^{4} x) d x \\ = \int_{}^{} (\sin^{2} x + \cos^{2} x) (\sin^{2} x - \cos^{2} x) d x \\ = \int_{}^{} (\sin^{2} x - \cos^{2} x) d x \\ = \int_{}^{} - \cos 2 x d x = - \frac{\sin 2 x}{2} + c \end{array}

Problem 5:

\begin{array}{l} \int_{}^{} \frac{x^{2} d x}{{(a + b x)}^{2}} = \end{array}

Solution:
Put a + bx = t ⇒ x = (t – a)/b and dx = dt/b

\begin{array}{l} I = {\int_{}^{} (\frac{t - a}{b})}^{2} \times \frac{1}{t^{2}} \frac{d t}{b} \\ = \frac{1}{b^{2}} \int_{}^{} (1 - \frac{2 a}{t} + a^{2} . t^{- 2}) d t \\ = \frac{1}{b^{2}} [t - 2 a \log t - \frac{a^{2}}{t}] \\ = \frac{1}{b^{2}} [x + \frac{a}{b} - \frac{2 a}{b} \log (a + b x) - \frac{a^{2}}{b} \frac{1}{(a + b x)}] \end{array}

Problem 6: Solve

\begin{array}{l} \int \frac{2 \cos x + 3 \sin x}{4 \cos x + 5 \sin x} d x \end{array}

Solution:

Problem of type

\begin{array}{l} \int \frac{a \cos x + b \sin x + p}{c \cos x + d \sin x + q} d x, \int \frac{a e^{x} + b e^{- x} + c}{d e^{x} + f e^{- x} + h} d x \end{array}

can be solved by

\begin{array}{l} N r = n D r + m D r^{'} \end{array}

Now,

Let 2 cos x + 3 sin x = a( 4 cos x + 5 sin x) + b(-4 sin x + 5 cos x)

Solving by comparing, we get

\begin{array}{l} a = \frac{23}{41} b = \frac{- 2}{41} \end{array}

\begin{array}{l} ∴ I = \int \frac{23}{41} - \frac{2}{41} (\frac{- 4 \sin x + 5 \cos x}{4 \cos x + 5 \sin x}) d x \end{array}

\begin{array}{l} = \frac{23}{41} x - \frac{2}{41} ℓ n | 4 \cos x + 5 \sin x | + c \end{array}

Problem 7: Find area between y² ≤ 4x, x² + y² ≥ 2x and x ≤ y + 2 in first quadrant.

Answer:

\begin{array}{l} A = \int_{0}^{{(\sqrt{3} + 1)}^{2}} \sqrt{4 x} d x - a r (s e m i c i r c l e) - a r (△ A B C) \end{array}

\begin{array}{l} = \sqrt{4} {[\frac{2 x^{3 / 2}}{3}]}_{0}^{{(\sqrt{3} + 1)}^{2}} - \frac{π}{2} - \frac{1}{2} {(\sqrt{3} + 1)}^{2} 2 (\sqrt{3} + 1) \end{array}

\begin{array}{l} = \frac{{(\sqrt{3} + 1)}^{3}}{3} - \frac{π}{2} \end{array}

Most Important Questions from Definite Integration for JEE Advanced

4,728

Frequently Asked Questions

Q1

What do you mean by integration in maths?

Integration is the process of finding the antiderivative of a function.

Q2

What is the integral of x?

The integral of x = (x²/2) + C, where C is the constant of integration.

Q3

What is the integral of sin x?

Integral of sin x =-cos x + C

Q4

Give two applications of integration.

Integration is used to find the area under a curve. It is also used to find the velocity and trajectory of a satellite.

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