The antiderivative of $g^{\prime }(x)$ 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})

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

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})

($$\begin{array}{l}\underset{0}{\overset{a}{\int }}f(x)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx\end{array}$$)

⇒ 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})

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

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})

($$\begin{array}{l}\text{Substitute}li{n}_1=t^2\to li{n}_1=t^2\end{array}$$ )

**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. ($$\begin{array}{l}-\underset{a}{\overset{b}{\int }}f(x)dx=\underset{b}{\overset{a}{\int }}f(x)dx\end{array}$$ )

$${\int }_a^{b}f(x),dx={\int }_a^{c}f(x),dx+{\int }_c^{b}f(x),dx$$

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

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

6. $${\int }_{-a}^{a}f(x)dx=\{\begin{array}{lll}0& \text{if }f(x)\text{ is odd}2{\int }_0^{a}f(x)dx& \text{if }f(x)\text{ is even}\end{array}$$

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\ $$\begin{array}{l}I=2\underset{0}{\overset{\pi }{\int }}{\sin }^{100}x{\cos }^{99}x\end{array}$$)

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

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

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

($$\begin{array}{l}=2\underset{0}{\overset{a}{\int }}f(2a-x)\text{if}f(2a-x)=f(x)\end{array}$$ )

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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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

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

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 $$Nr=nDr+mD{r}^{\prime }$$ 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})

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

Problem 7: Find the area of the region bounded by the curves $y^2\le 4x$, $x^2+y^2\ge 2x$, and $x\le y+2$ 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