Important Integration Formulas for Jee

In calculus, integration plays a crucial role, especially in JEE exams. Mastering the fundamental integration formulas ensures that students can solve complex problems efficiently. This blog covers a list of all integral formulas that are essential for JEE, along with their examples to help you understand their applications better.

1.0Introduction to Integration

Integration is the inverse process of differentiation. If a F(x) is the integral of f(x), then F'(x) = f(x). It helps in finding the area under a curve, volumes, and solving differential equations, which are important in physics and mathematics.

2.0List of All Integral Formulas

Standard Formulae

1. $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C;n\ne -1$
2. $\int \frac{dx}{ax+b}=\frac{1}{a}\ln |ax+b|+C$
3. $\int e^{ax+b}dx=\frac{1}{a}{e}^{ax+b}+C$
4. $\int a^{px+q}dx=\frac{1}{p}\frac{a^{px+q}}{\ln a}+C,(a>0)$
5. $\int \sin (ax+b)dx=-\frac{1}{a}\cos (ax+b)+C$
6. $\int \cos (ax+b)dx=\frac{1}{a}\sin (ax+b)+C$
7. $\int \tan (ax+b)dx=\frac{1}{a}\ln |\sec (ax+b)|+C$
8. $\int \cot (ax+b)dx=\frac{1}{a}\ln |\sin (ax+b)|+C$
9. $\int {\sec }^2(ax+b)dx=\frac{1}{a}\tan (ax+b)+C$
10. $\int {\mathrm{cosec}}^2(ax+b)dx=-\frac{1}{a}\cot (ax+b)+C$
11. $\int \mathrm{cosec}(ax+b)\cdot \cot (ax+b)dx=-\frac{1}{a}\mathrm{cosec}(ax+b)+C$
12. $\int \sec (\mathrm{ax}+\mathrm{b})\cdot \tan (\mathrm{ax}+\mathrm{b})\mathrm{dx}=\frac{1}{a}\sec (ax+b)+C$
13. $\int \sec xdx=\ln |\sec x+\tan x|+C=\ln \left|\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\right|+C$
14. $\int \mathrm{cosec}xdx=\ln |\mathrm{cosec}x-\cot x|+C=\ln \left|\tan \frac{x}{2}\right|+C=-\ln |\mathrm{cosec}x+\cot x|+C$
15. $\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+C$
16. $\int \frac{dx}{a^2+x^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C$
17. $\int \frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}{\sec }^{-1}\frac{x}{a}+C$
18. $\int \frac{dx}{\sqrt{x^2+a^2}}=\ln \left[x+\sqrt{x^2+a^2}\right]+\mathrm{C}$
19. $\int \frac{dx}{\sqrt{x^2-a^2}}=\ln \left[x+\sqrt{x^2-a^2}\right]+\mathrm{C}$
20. $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+C$
21. $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln \left|\frac{x-a}{x+a}\right|+C$
22. $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$
23. $\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln \left(x+\sqrt{x^2+a^2}\right)+C$
24. $\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln \left(x+\sqrt{x^2-a^2}\right)+C$
25. $\int e^{ax}\cdot \sin bxdx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)+C=\frac{e^{ax}}{\sqrt{a^2+b^2}}\sin \left(bx-{\tan }^{-1}\frac{b}{a}\right)+C$
26. $\int e^{ax}\cdot \cos bxdx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+C=\frac{e^{ax}}{\sqrt{a^2+b^2}}\cos \left(bx-{\tan }^{-1}\frac{b}{a}\right)+C$

Methods of Integration

Substitution or Change of Independent Variable

If $\varphi (x)$ is a continuous differentiable function, then to evaluate integrals of the form $\int f(\varphi (x)){\varphi }^{\prime }(x)dx$, we substitute $\varphi (\mathrm{x})=\mathrm{t}$ and ${\varphi }^{\prime }(\mathrm{x})dx=dt$ .

Hence $I=\int f(\varphi (x)){\varphi }^{\prime }(x)dx$ reduces to $\int f(t)dt$.

Integral of the form 

• $\int \frac{dx}{a{x}^2+bx+c},\int \frac{dx}{\sqrt{a{x}^2+bx+c}}$

Express a $x^2+bx+c$ in the form of a perfect square & then apply the standard results.

• $\int \frac{px+q}{a{x}^2+bx+c}dx,\int \frac{px+q}{\sqrt{a{x}^2+bx+c}}dx$

Express p x+q=l (differential coefficient of denominator ) +m .

Integration by Part: 

$\int u\cdot vdx=u\int vdx-\int \left[\frac{du}{dx}\cdot \int vdx\right]dx$

where u & v are differentiable functions and are commonly designated as first & second function respectively. 

Note:  While using integration by parts, choose u & v such that 

(i) $\int vdx$ &

(ii) $\int {\left[\frac{du}{dx}\cdot \int vdx\right]}_{dx}$  are simple to integrate.

This is generally obtained by choosing the first function refers to the function that appears first in the acronym ILATE, which stands for:

• I - Inverse function
• L - Logarithmic function
• A - Algebraic function
• T - Trigonometric function
• E - Exponential function.

This order is used to determine the sequence in which functions should be integrated when applying integration techniques.

Let

$\begin{array}{rl}& \mathrm{I}=\int f(x)\cdot g(x)dx\\ & \text{ I II }\end{array}$

$=f(x)\cdot \int g(x)dx-\int \left(f^{\prime }(x)\right)\left(\int g(x)dx\right)dx$

$=1^{st}function\times integralof{2}^{nd}-\int \left(\text{ diff. coeff. of }{\mathrm{I}}^{st}\right)\times \left(\text{ integral of }{2}^{nd}\right)dx$

Two Classic Integrands

(a) $\int e^x\left(f(x)+f^{\prime }(x)\right)dx=e^{x}f(x)+c$

(b) $\int \left(f(x)+x{f}^{\prime }(x)\right)dx=xf(x)+c$

Integrals of the Type:

(i) $\int e^{ax}\cdot \sin bxdx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)+c$

(ii)$\int e^{ax}\cdot \cos bxdx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+c$

Integration of Rational Function                                   

(i) A rational function is defined as the quotient of two polynomials, expressed in the form

$\frac{P(x)}{Q(x)}$​, where P(x) and Q(x) are polynomials in x and Q(x) ≠ 0. If the degree of P(x) is less than the degree of Q(x), the rational function is classified as proper. Conversely, if the degree of P(x) is greater than or equal to the degree of Q(x), it is termed improper. An improper rational function can be transformed into a proper rational function through the process of long division.

Thus, if a rational function is improper, it can be expressed as:

$\frac{P(x)}{Q(x)}=H(x)+\frac{R(x)}{Q(x)}$​,

where H(x) is a polynomial and $\frac{R(x)}{Q(x)}$​ is a proper rational function. To simplify the integration of such functions, we use a technique known as partial fraction decomposition, which breaks the integrand into a sum of simpler rational functions. Once decomposed, the integral can be evaluated easily using standard integration methods.

Manipulating Integrands

(i) $\int \frac{dx}{x\left(x^n+1\right)}$, $n\in N$, take $x^n$ common & put $1+x^{-n}=t$.

(ii) $\int \frac{dx}{x^2{\left(x^n+1\right)}^{(n-1)/n}},n\in N$,  take $x^n$ common & put $1+x^{-n}=t^n$

(iii) $\int \frac{dx}{x^n{\left(1+x^n\right)}^{1/n}}$, take $x^n$ common and put $1+x^{-n}=t^n$ . 

Integral of the Form: 

• $\int \frac{dx}{a+b{\sin }^{2}x}$ OR $\int \frac{dx}{a+b{\cos }^{2}x}$

OR $\int \frac{dx}{a{\sin }^{2}x+b\sin x\cos x+c{\cos }^{2}x}$ 

Divide  $N^r$ &  $D^r$ by x  &  put $\tan x=t$.

• $\int \frac{dx}{a+b\sin x}OR\int \frac{dx}{a+b\cos x}OR\int \frac{dx}{a+b\sin x+c\cos x}$

Convert sines and cosines into their respective tangents of half the angles, put $\tan \frac{x}{2}=t$_.

• Integrals of the form $\int \frac{x^2+1}{x^4+K{x}^2+1}dxOR\int \frac{x^2-1}{x^4+K{x}^2+1}dx$ , where K is any constant.

Divide $N^r$ & $D^r$ by $x^2$ then put $x-\frac{1}{x}=tORx+\frac{1}{x}=t$ respectively & proceed.

Integration of Irrational functions:

(i) $\int \frac{dx}{(ax+b)\sqrt{px+q}}and\int \frac{dx}{\left(a{x}^2+bx+c\right)\sqrt{px+q}}\text{; put }px+q=t^2$

(ii) $\int \frac{dx}{(ax+b)\sqrt{p{x}^2+qx+r}};\int \frac{dx}{\left(a{x}^2+b\right)\sqrt{p{x}^2+q}},putx=\frac{1}{t}$

3.0Definite Integration

A definite integral is denoted by ${\int }_a^{b}f(x)dx$ which represents the algebraic area bounded by the curve y = f(x), the ordinates x = a, x = b and the x-axis.

Properties of Definite Integral

(a) ${\int }_a^{b}f(x)dx={\int }_a^{b}f(t)dt\Rightarrow {\int }_a^{b}f(x)dx$ does not depend upon x. It is a numerical quantity.

_(b) ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$

_(c) ${\int }_a^{b}f(x)dx=-{\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$_, where c may lie inside or outside the interval [a, b]. This property to be used when f is piecewise continuous in (a, b). 

(d) ${\int }_{-a}^{a}f(x)dx={\int }_0^a[f(x)+f(-x)]dx=[\begin{array}{cl}0& \text{; if }f(x)\text{ is an odd function }\\ 2{\int }_0^{a}f(x)dx& \text{; if }f(x)\text{ is an even function }\end{array}$

(e) ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx,Inparticular{\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

(f) ${\int }_0^{2a}f(x)dx={\int }_0^{a}f(x)dx+{\int }_0^{a}f(2a-x)dx=[\begin{array}{cc}2{\int }_0^{a}f(x)dx& \text{; if }f(2a-x)=f(x)\\ 0& \text{; if }f(2a-x)=-f(x)\end{array}$

(g) ${\int }_0^{nT}f(x)dx=n{\int }_0^{T}f(x)dx,(n\in I)$; where ‘T’ is the period of the function i.e. f(T + x) = f(x)

Note that : ${\int }_x^{T+x}f(t)dt$ will be independent of x and equal to ${\int }_0^{T}f(t)dt$ 

(h) ${\int }_{a+nT}^{b+nT}f(x)dx={\int }_a^{b}f(x)dx$ where f(x) is periodic with period T & n ∈ I.

(i) ${\int }_{ma}^{na}f(x)dx=(n-m){\int }_0^{a}f(x)dx,(n,m\in I)if(x)$ is periodic with period ‘a’.

Some standard results

(i) ${\int }_0^{\pi /2}\frac{{\sin }^{\mathrm{n}}\mathrm{x}}{{\sin }^{\mathrm{n}}\mathrm{x}+{\cos }^{\mathrm{n}}\mathrm{x}}\mathrm{dx}={\int }_0^{\pi /2}\frac{{\cos }^{\mathrm{n}}\mathrm{x}}{{\cos }^{\mathrm{n}}\mathrm{x}+{\sin }^{\mathrm{n}}\mathrm{x}}\mathrm{dx}=\frac{\pi }{4}$

(ii) ${\int }_0^{\pi /2}\log \tan xdx=0$

(iii) ${\int }_0^{\pi /4}\log (1+\tan x)dx=\frac{\pi }{8}\ln 2$

(iv) ${\int }_0^{\pi /2}\ln (\sin x)dx={\int }_0^{\pi /2}\ln (\cos x)dx=\frac{-\pi }{2}\ln 2$

(v) ${\int }_a^b\frac{f(x)}{f(x)+f(a+b-x)}dx=\frac{b-a}{2}$

(vi) ${\int }_{\mathrm{a}}^{\mathrm{b}}\{\mathrm{x}\}\mathrm{dx}=\frac{\mathrm{b}-\mathrm{a}}{2},\mathrm{a},\mathrm{b}\in \mathrm{I}$

(vii) ${\int }_{\mathrm{a}}^{\mathrm{b}}\frac{|\mathrm{x}|}{\mathrm{x}}\mathrm{dx}=|\mathrm{b}|-|\mathrm{a}|$

Walli’s Theorem

(a) ${\int }_0^{\pi /2}{\sin }^{n}xdx={\int }_0^{\pi /2}{\cos }^{n}xdx=\frac{(n-1)(n-3)\dots ..(1\text{ or }2)}{n(n-2)\dots ..(1\text{ or }2)}K$

where $K=\{\begin{array}{ll}\pi /2& \text{ if }n\text{ is even }\\ 1& \text{ if }n\text{ is odd }\end{array}$

(b) ${\int }_0^{\pi /2}{\sin }^{n}x\cdot {\cos }^{m}xdx=\frac{[(n-1)(n-3)(n-5)\dots ..1\text{ or }2][(m-1)(m-3)\dots .\dots 1\text{ or }2]}{(m+n)(m+n-2)(m+n-4)\dots ..1\text{ or }2}K$

Where $K=\{\begin{array}{l}\frac{\pi }{2}\text{ if both }m\text{ and }n\text{ are even }(m,n\in N)\\ 1\text{ otherwise }\end{array}$

Leibnitz Theorem

If $\mathrm{F}(\mathrm{x})={\int }_{\mathrm{g}(\mathrm{x})}^{\mathrm{h}(\mathrm{x})}\mathrm{f}(\mathrm{t})\mathrm{dt}then{F}^{\prime }(x)=h^{\prime }(x)f(h(x))-g^{\prime }(x)f(g(x))$

Where h(x) and g(x) are differentiable function of x.

Estimation of D.I. & General Inequality

If f(x) is continuous in [a, b] and it’s range in this interval is [m, M], then 

$m(b-a)\le {\int }_a^{b}f(x)dx\le M(b-a)$

Definite Integral as Limit of a Sum:

${\int }_a^{b}f(x)dx={\mathrm{Lim}}_{n\to \infty }h[f(a)+f(a+h)+f(a+2h)+\dots .+f(a+\overline{n-1}h)]$

${\mathrm{Lim}}_{n\to \infty }h{\sum }_{r=0}^{n-1}f(a+rh)$, where b – a = nh

If a = 0 & b = 1 then, ${\mathrm{Lim}}_{n\to \infty }h{\sum }_{r=0}^{n-1}f(rh)={\int }_0^{1}f(x)dx;wherenh=1Or{\mathrm{Lim}}_{n\to \infty }\left(\frac{1}{n}\right){\sum }_{r=0}^{n-1}f\left(\frac{r}{n}\right)={\int }_0^{1}f(x)dx$.

4.0Solved Example of important Integral Formula

Example 1: Evaluate $\int \frac{\left(x^2-1\right)dx}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(x+\frac{1}{x}\right)}$

Solution:

The given integral can be written as

$\mathrm{I}=\int \frac{\left(1-\frac{1}{x^2}\right)dx}{\left[{\left(x+\frac{1}{x}\right)}^2+1\right]{\tan }^{-1}\left(x+\frac{1}{x}\right)}$

Let $\left(x+\frac{1}{x}\right)=t$. Differentiating we get  $\left(1-\frac{1}{x^2}\right)dx=dt$

Hence $I=\int \frac{dt}{\left(t^2+1\right){\tan }^{-1}t}$

Now make one more substitution ${\tan }^{-1}t=u$. Then $\frac{dt}{t^2+1}=duandI=\int \frac{du}{u}=\ln |u|+C$

Returning to t, and then to x, we have

Ans. $\mathrm{I}=\ln \left|{\tan }^{-1}t\right|+C=\ln \left|{\tan }^{-1}\left(x+\frac{1}{x}\right)\right|+C$

Example 2: Evaluate $\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\cdot \frac{1}{x}dx$ 

Solution:

Put $x={\cos }^2\theta \Rightarrow dx=-2\sin \theta \cos \theta d\theta$

$\Rightarrow I=\int \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\cdot \frac{1}{{\cos }^2\theta }(-2\sin \theta \cos \theta )d\theta =-\int 2\tan \frac{\theta }{2}\tan \theta d\theta$

$=-4\int \frac{{\sin }^2(\theta /2)}{\cos \theta }d\theta =-2\int \frac{1-\cos \theta }{\cos \theta }d\theta =-2\ln |\sec \theta +\tan \theta |+2\theta +C$

$=-2\ln \left|\frac{1+\sqrt{1-x}}{\sqrt{x}}\right|+2{\cos }^{-1}\sqrt{x}+C$

Example 3: Evaluate $\int {\sin }^{2}x{\cos }^{4}xdx$

Solution:

$\int {\sin }^{2}x{\cos }^{4}xdx=\int \left(\frac{1-\cos 2x}{2}\right){\left(\frac{\cos 2x+1}{2}\right)}^{2}dx$

$=\int \frac{1}{8}(1-\cos 2x)\left({\cos }^{2}2x+2\cos 2x+1\right)dx$

$=\frac{1}{8}\int \left({\cos }^{2}2x+2\cos 2x+1-{\cos }^{3}2x-2{\cos }^{2}2x-\cos 2x\right)dx$

$=\frac{1}{8}\int \left(-{\cos }^{3}2x-{\cos }^{2}2x+\cos 2x+1\right)dx$

$=-\frac{1}{8}\int \left(\frac{\cos 6x+3\cos 2x}{4}+\frac{1+\cos 4x}{2}-\cos 2x-1\right)dx$

$=-\frac{1}{32}\left[\frac{\sin 6x}{6}+\frac{3\sin 2x}{2}\right]-\frac{1}{16}x-\frac{\sin 4x}{64}+\frac{\sin 2x}{16}+\frac{x}{8}+C$

$=-\frac{\sin 6x}{192}-\frac{\sin 4x}{64}+\frac{1}{64}\sin 2x+\frac{x}{16}+C$

Example 4: Evaluate $\int \frac{x^4}{(x+2)\left(x^2+1\right)}dx$

Solution:

$\frac{x^4}{(x+2)\left(x^2+1\right)}=(x-2)+\frac{3x^2+4}{(x+2)\left(x^2+1\right)}$

Now,

$\frac{3x^2+4}{(x+2)\left(x^2+1\right)}=\frac{16}{5(x+2)}+\frac{-\frac{1}{5}x+\frac{2}{5}}{x^2+1}$

_So, $\frac{x^4}{(x+2)\left(x^2+1\right)}=x-2+\frac{16}{5(x+2)}+\frac{-\frac{1}{5}x+\frac{2}{5}}{x^2+1}$

Now, $\int (x-2)+\frac{16}{5(x+2)}+\frac{-\frac{1}{5}x+\frac{2}{5}}{x^2+1})dx$

Ans. $=\frac{x^2}{2}-2x+\frac{2}{5}{\tan }^{-1}x+\frac{16}{5}\ln |x+2|-\frac{1}{10}\ln \left(x^2+1\right)+C$

Example 5: Evaluate

$\int \frac{{\cos }^{4}xdx}{{\sin }^{3}x{\left\{{\sin }^{5}x+{\cos }^{5}x\right\}}^{\frac{3}{5}}}$

Solution:

$I=\int \frac{{\cos }^{4}x}{{\sin }^{3}x{\left\{{\sin }^{5}x+{\cos }^{5}x\right\}}^{\frac{3}{5}}}dx$

$=\int \frac{{\cos }^{4}x}{{\sin }^{6}x{\left\{1+{\cot }^{5}x\right\}}^{\frac{3}{5}}}dx$

$=\int \frac{{\cot }^{4}x{\mathrm{cosec}}^{2}xdx}{{\left(1+{\cot }^{5}x\right)}^{3/5}}$

$\text{ Put }1+{\cot }^{5}x=t\Rightarrow 5{\cot }^{4}x{\mathrm{coses}}^{2}xdx=-dt$

Ans. $=-\frac{1}{5}\int \frac{dt}{t^{3/5}}=-\frac{1}{2}{t}^{2/5}+C=-\frac{1}{2}{\left(1+{\cot }^{5}x\right)}^{2/5}+C$

Example 6: Evaluate ${\int }_0^2\frac{dx}{\left(17+8x-4x^2\right)\left[e^{6(1-x)}+1\right]}$

Solution:

Let $I={\int }_0^2\frac{dx}{\left(17+8x-4x^2\right)\left[e^{6(1-x)}+1\right]}$

Also $I=\frac{dx}{\left(17+8x-4x^2\right)\left[e^{-6(1-x)}+1\right]}\left[{\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx\right]$

Adding, we get 

$2\mathrm{I}={\int }_0^2\frac{1}{17+8x-4x^2}\left(\frac{1}{e^{6(1-x)}+1}+\frac{1}{e^{-6(1-x)}+1}\right)dx$

$={\int }_0^2\frac{1}{17+8x-4x^2}dx=-\frac{1}{4}{\int }_0^2\frac{dx}{x^2-2x-17/4}$

$=-\frac{1}{4}{\int }_0^2\frac{dx}{(x-1{)}^2-21/4}=-\frac{1}{4}\times \frac{1}{2\times \frac{\sqrt{21}}{2}}{\left[\log \left|\frac{x-1-\frac{\sqrt{21}}{2}}{x-1+\frac{\sqrt{21}}{2}}\right|\right]}_0^2$

$=-\frac{1}{4\sqrt{21}}{\left[\log \left|\frac{\mid 2x-2-\sqrt{21}}{2x-2+\sqrt{21}}\right|\right]}_0^2$

$\Rightarrow I=-\frac{1}{8\sqrt{21}}\left[\log \left|\frac{2-\sqrt{21}}{2+\sqrt{21}}\right|-\log \left|\frac{2+\sqrt{21}}{\sqrt{21}-2}\right|\right]$

$=-\frac{1}{4\sqrt{21}}\left[\log \left|\frac{\sqrt{21}-2}{2+\sqrt{21}}\right|\right]$

5.0Practice Question for Integration for JEE

1. $\int \left[\frac{{\cot }^{2}2x-1}{2\cot 2x}-\cos 8x\cot 4x\right]dx$
2. $\int \left(\frac{\frac{x}{x+1}-\ln (x+1)}{x(\ln (x+1))}\right)dx$
3. $\int \frac{dx}{{\cos }^{2}x\sqrt{1+\tan x}}$
4. $\int \frac{dx}{\cot \frac{x}{2}\cdot \cot \frac{x}{3}\cdot \cot \frac{x}{6}}$
5. $\int \frac{x^2{\tan }^{-1}x}{1+x^2}dx$