The antiderivative of $g’(x)$ with respect to $dx$ is found through the process of integration, and is given by:

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

The two types of integrals include:

• Indefinite Integrals
• Definite Integrals

Definite Integral: An integral with specified upper and lower limits, without the constant of integration.

Indefinite integral: An integral without limits and an arbitrary constant added.

This article provides a comprehensive overview of standard integrals, their properties, important formulas, and examples of integration, which will help students gain a deeper understanding of the topic.

Standard Integrals

Integrals of Rational and Irrational Functions

(\begin{array}{l} \int x^n , dx = \frac{x^{n+1}}{n+1} + C, \quad n \ne 1 \ \int \frac{1}{x} , dx = \ln |x| + C \ \int c , dx = c \cdot x + C \ \int x , dx = \frac{x^2}{2} + C \ \int x^2 , dx = \frac{x^3}{3} + C \ \int \frac{1}{x^2} , dx = -\frac{1}{x} + C \\end{array})

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

Integrals of Trigonometric Functions

(\begin{array}{l} \int \sin x,dx = -\cos x + C \ \int \cos x,dx = \sin x + C \ \int \tan x,dx = \ln|\sec x| + C \ \int \sec x,dx = \ln|\tan x + \sec x | + C \\end{array})

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

Integrals of Exponential and Logarithmic Functions

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

Properties of Integration

Property 1: (\int\limits_{a}^{a} f(x),dx = 0)

Property 2: (\int\limits_{a}^{b}{f(x),dx=-\int\limits_{b}^{a}{f(x),dx}})

Property 3: (\int\limits_{a}^{b}{f(x) , dx} = \int\limits_{a}^{b}{f(t) , dt})

Property 4: (\int\limits_{a}^{b}{f(x),dx} = \int\limits_{a}^{c}{f(x),dx} + \int\limits_{c}^{b}{f(x),dx})

Property 5: (\int\limits_{a}^{b}{f(x)dx} = \int\limits_{a}^{b}{f(a+b-x)dx})

(\begin{array}{l}\int\limits_{0}^{a}{f(x)dx} = \int\limits_{0}^{a}{f(a-x)dx}\end{array})

⇒ Also Read Definite and Indefinite Integration

Useful Formulas

* (\int{{e}^{ax}}\sin bx=\frac{{e}^{ax}}{{a}^{2}+{b}^{2}}\left[ a\sin bx-b\cos bx \right])

* (\int{{{e}^{ax}}\cos bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ a\cos bx+b\sin bx \right]}\)

*(\int{{e}^{x}}\left( f(x)+f’(x) \right) = {e}^{x}f(x))

Illustration:

(\int{{e}^{x}}(\sin x+\cos x)dx={{e}^{x}}\sin x+c)

(\int{{e}^{x}(lnx+\frac{1}{x})dx={{e}^{x}}lnx+c)

Integrating Trigonometric Functions

Type 1: (\int{{{\sin }^{m}}x{{\cos }^{n}}xdx})

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

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

If m and n are rational, then put tan x = t

If both are even, then use the reduction method

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

t = sin(x)

(\int{{{t}^{-5}}dt})

(\begin{array}{l}=\frac{-1}{5\sin^5x}+\frac{1}{3\sin^3x}+c\end{array})

Type 2: (\int{\frac{dx}{a\cos x + b\sin x + c}})

t = \tan\left(\frac{x}{2}\right)

Illustration

(\begin{array}{l}\int{\frac{dx}{2+\sin x}} = \int{\frac{d(2\tan\left(\frac{x}{2}\right))}{2+\sin x}} \end{array} )

(\begin{array}{l}\Rightarrow \int{\frac{d(2\tan\left(\frac{x}{2}\right))}{2+\sin x}} = \int{2\frac{dt}{1+t^2}} \end{array} )

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

(\frac{d}{dt}\left( {{t}^{2}} \right)=2t)

(\begin{array}{l}=\int{\frac{dt}{{{t}^{2}}+t+1}}\end{array})

(\frac{2}{\sqrt{3}}\arctan\left(\frac{2t+1}{\sqrt{3}}\right))

(\begin{array}{l}=\frac{2}{\sqrt{3}}\arctan\left(\frac{2\tan\frac{x}{2}+1}{\sqrt{3}}\right)+c\end{array})

Substitutions for Irrational Functions

Form 1: (\displaystyle \int \sqrt{Quadratic} \ dx )

(\begin{array}{l}\text{Substitute}\ m=Qudratic,\ n=Linear\end{array})

Form 2: (\int{\frac{dx}{\sqrt{l_{1}in}}},,\int{\frac{l_{1}in}{\sqrt{l_{1}in}}}dx,,\int{\frac{\sqrt{l_{1}in}}{l_{1}in}dx})

(\begin{array}{l}\text{Substitute}\ li{{n}_{1}}={{t}^{2}} \ \rightarrow \ li{{n}_{1}}={{t}^{2}} \end{array} )

**Form 3:** (\int{\frac{1}{\sqrt{Qua}}dx})

Substitute $$lin = \frac{1}{t}$$

Form 4: (\int{\frac{dx}{\left( a{{x}^{2}}+b \right)\sqrt{\left( {{x}^{2}}+d \right)}})

Substitute x = $\frac{1}{t}$ and then $u^2$ for $a t^2 + b$

Integration Formulas

1. (\int\limits_{a}^{b}{f(x),dx} = \int\limits_{a}^{b}{f(t),dt})

2. (\begin{array}{l}-\int\limits_{a}^{b}{f(x)dx=}\int\limits_{b}^{a}{f(x)dx}\end{array} )

$$\int_a^b f(x) , dx = \int_a^c f(x) , dx + \int_c^b f(x) , dx$$

4. (\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(a+b-x)dx}})

5. $$\int\limits_{0}^{2a}{f(x)dx}=\int\limits_{0}^{a}{f(x)dx}+\int\limits_{0}^{a}{f(2a-x)dx}$$ $$=0 \quad \text{if} \quad f(2a-x)=-f(x)$$ and $$=2\int\limits_{0}^{a}{f(x)} \quad \text{if} \quad f(2a-x)=f(x)$$

6. $$\int_{-a}^{a}{f(x)dx=\begin{cases} 0 & \text{if } f(x) \text{ is odd} \ 2\int_{0}^{a}{f(x)dx} & \text{if } f(x) \text{ is even} \end{cases}}$$

Problems on Integration

Illustration:

$$\int_{0}^{2}{{x}^{2}\left[ x \right]dx}=\int_{0}^{1}{{x}^{2}\left[ x \right]dx}+\int_{1}^{2}{{x}^{2}\left[ x \right]dx}$$

(\int\limits_{0}^{1}{x^2,dx} + \int\limits_{1}^{2}{x^2,dx})

$\int_{1}^{2}\frac{x^3}{3}dx = 0$

(\frac{8-1}{3} = \frac{7}{3})

Illustration:

(\int\limits_{{\pi }/{6}}^{{\pi }/{3}}{\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx})

(\begin{array}{l}I = \int{\frac{\sqrt{\cos (\frac{\pi}{2} – x)}}{\sqrt{\sin (\frac{\pi}{2} – x)}+\sqrt{\cos (\frac{\pi}{2} – x)}}dx}\end{array})

(\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}{\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}},dx} = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}{1,dx} = \frac{\pi}{6})

(\begin{array}{l}I=\frac{\pi }{12}\end{array})

Illustration:

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

Here f(2π - x) = f(x)

(Or\ \begin{array}{l}I=2\int\limits_{0}^{\pi}{{{\sin }^{100}}x{{\cos }^{99}}x}\end{array})

(\int\limits_{0}^{\pi }{{{\sin }^{100}}\left( \pi -x \right){{\cos }^{99}}\left( \pi -x \right)},dx = 2)

-I = I

I = 0

Illustration:

(\int\limits_{-5}^{5}{{{x}^{3}}\ \text{d}x=0} \ \text{as}\ f(x)=x^3 \ \text{is\ an\ odd\ function})

Leibnitz’s Rule

(\frac{d}{dx}\int\limits_{u(x)}^{v(x)}{f(t)dt} = f(v(x))\frac{dv(x)}{dx} - f(u(x))u’(x))

Practice Problems

Problem 1. (\begin{array}{l}\text{If}\ \int\limits_{x^2}^{x^3}{\frac{1}{\log t}dt=y},\ \text{find}\ \frac{dy}{dx}\end{array})

(\frac{dy}{dx}=x\left( x-1 \right){{\left( \log x \right)}^{-1}})

Problem 2. If $$\int\limits_{\sin x}^{1}{{{t}^{2}}f\left( t \right)dt=1-\sin x.$$ where $x \in (0, \frac{\pi}{2})$, find $f(\frac{1}{\sqrt{3}})$.

Problem 3. (\displaystyle \lim_{x \to 2} \int_{6}^{f(x)}\frac{4t^{3}}{x-2}dt = 18.)

Problem 4. (\displaystyle \lim_{x\to \infty }\frac{\int\limits_{0}^{x}{{{e}^{{{x}^{2}}}}dx}}{\int\limits_{0}^{x}{{{e}^{2{{x}^{2}}}}dx}}=0)

Integration by Parts

(\int{uv,dx} = u\int{vdx} - \int{u’\left( \int{vdx} \right)},dx)

Illustration:

Q. (\int{\ln,x,dx} = \int{\ln,x.1,dx})

(\begin{array}{l}x,\ell n,x - \int{x,dx - \int{\frac{1}{x},dx}}\end{array})

(\begin{array}{l}=\ell n,x-1\end{array})

Q. (\int x,{{e}^{x}}dx = x\int{{{e}^{x}}dx - \int{{{\left( 1 \right)}}\left( \int{{{e}^{x}}dx} \right)}dx} )

(\int x{{e}^{x}}dx = x{{e}^{x}} - \int{{{e}^{x}}dx})

\(\frac{d}{dx} \left( xe^x - e^x \right) \)

Integration of Irrational Algebraic Functions

Type $\int{\frac{dx}{{{\left( ax+b \right)}^{k}}\sqrt{px+q}}};$

Q. (\int{\frac{x}{\left( x-3 \right)\sqrt{x+1}}dx})

x + 1 = t2 $\Rightarrow$ x = t2 - 1

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

\(\int{2}+\frac{3}{{{t}^{2}}-4}dt\)

(\begin{array}{l}2t + \frac{3}{2} \ln \left| \frac{t-2}{t+2} \right| + c\end{array})

(\begin{array}{l}2\sqrt{x+1} + \frac{3}{2}\ln\left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right| + c\end{array})

(\int\limits_{0}^{2a}{f\left( x \right)dx} = \int\limits_{0}^{a}{f\left( x \right)dx} + \int\limits_{0}^{a}{f\left( 2a-x \right)dx})

f(2a - x) = -f(x) \Rightarrow 0

(\begin{array}{l}=2\int\limits_{0}^{a}{f\left( 2a - x \right)}\ \text{if}\ f(2a – x) = f(x)\end{array} )

Optimizing Area for Maximum and Minimum Values

Illustration:

f(x) = x^2 + 2

I love to listen to music!

Answer: I enjoy listening to music!

(\int\limits_{2}^{\alpha }{\left( {{x}^{2}}+2-f\left( x \right) \right)dx} ={{\alpha }^{3}}-4{{\alpha }^{2}}+8 )

Differentiating the Labniz Equation

(\begin{array}{l}3{{\alpha }^{2}}-8\alpha - {{\alpha }^{2}} - 2 + f\left( \alpha \right) = 0\end{array})

$f(x) = -2x^2 + 8x + 2$

Important JEE Main Questions for Integrations

Definite Integration JEE Questions

![Definite Integration]()

Indefinite Integration JEE Questions

![Indefinite Integration]()

Solved Problems on Integration

Problem 1: (\int_{a}^{b}{\frac{dx}{\cos (x-a)\cos (x-b)}})

Given:

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Solution:

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

Problem 2: (\int_{}^{}\frac{dx}{\sqrt{x+a}+\sqrt{x+b}})

Given:

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Solution:

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

Problem 3: (\int_{}^{}{\frac{x^3-x-2}{(1-x^2)} \ dx})

Given:

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Solution:

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$$\int_{}^{}{\frac{x^3-x-2}{(1-x^2)},dx} = \int_{}^{}{\frac{-x(1-x^2)}{(1-x^2)},dx-\int_{}^{}{\frac{2}{1-x^2},dx}}$$ $$=-\int_{}^{}{x,dx}-2\int_{}^{}{\frac{1}{1-x^2},dx}$$ $$=\frac{-x^2}{2}+\log \left(\frac{x-1}{x+1}\right)+c.$$

Problem 4: (\int_{}^{}\frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}\ dx)

Given:

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Solution:

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

Problem 5: (\int_{}^{}{\frac{x^2 , dx}{(a+bx)^2}} = )

Given:

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Solution:

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x = (t - a) / b
dx = dt/b

(\begin{array}{l}I= \frac{1}{{{b}^{2}}}\left[ x+\frac{a}{b}-\frac{2a}{b}\log (a+bx)-\frac{{{a}^{2}}}{b}\frac{1}{(a+bx)} \right]\end{array})

Problem 6: Solve (\int{\frac{2\cos x+3\sin x}{4\cos x+5\sin x}},dx)

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Solution:

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Problem of type $$\int{\frac{a\cos x+b\sin x+p}{c\cos x+d\sin x+q}},dx,\int{\frac{a{{e}^{x}}+b{{e}^{-x}}+c}{d{{e}^{x}}+f{{e}^{-x}}+h}},dx$$ can be solved by $$Nr=nDr+mDr’$$ she said

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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} )

(\therefore I=\int{\frac{25}{41}\left( \frac{-4\sin x+5\cos x}{4\cos x+5\sin x} \right)dx})

(\begin{array}{l} \frac{23}{41}x-\frac{2}{41}\ln\left| 4\cos x+5\sin x \right| + c \end{array})

Problem 7: Find the area of the region bounded by the curves $y^2 \leq 4x$, $x^2 + y^2 \geq 2x$, and $x \leq y + 2$ in the first quadrant.

Given:

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Answer:

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(\int\limits_{0}^{{{\left( \sqrt{3}+1 \right)}^{2}}}{\sqrt{4x},,dx-} ar(semicircle) + ar(\triangle ABC))

(\frac{2{{x}^{3/2}}}{3}_{0}^{{{\left( \sqrt{3}+1 \right)}^{2}}}-\frac{1}{2}{{\left( \sqrt{3}+1 \right)}^{2}}2\left( \sqrt{3}+1 \right)-\sqrt{4}\left( \frac{\pi }{2} \right))

(\frac{{{\left( \sqrt{3}+1 \right)}^{3}}{3} - \frac{\pi}{2})

Most Important Questions from Definite Integration for JEE Advanced

Frequently Asked Questions

Integration in Maths refers to the process of calculating the area under a curve by breaking it down into smaller sections and adding them together.

The process of finding the antiderivative of a function is known as Integration.

The integral of x is $\int x ,dx = \frac{x^2}{2} + C$

Integral of x = $\int x \; dx = \frac{x^2}{2} + C$, where $C$ is the constant of integration.

The integral of sin x is -cos x + C, where C is an arbitrary constant.

Integral of $\sin x = -\cos x + C$

1. Finding the area under a curve
2. Computing the volume of a solid of revolution

Integration is used to:

• Find the area under a curve
• Find the velocity of a satellite
• Find the trajectory of a satellite