Indefinite Integration

If f & F are function of x such that F'(x) = f(x) then the function F is called a Primitive or Antiderivative or Integral of f(x) w.r.t. x and is written symbolically as

$\int f(x)dx=F(x)+C\iff \frac{d}{dx}\{F(x)+C\}=f(x)$ where C is called the constant of integration.

1.0Indefinite Integration Definition

Indefinite Integration, also known as anti-differentiation, is a fundamental concept in calculus. It involves finding a function whose derivative is a given function. If we have a function f(x), indefinite integration aims to determine another function F(x) such that when we differentiate F(x), we get f(x). 

In simpler terms, indefinite integration is the reverse process of differentiation. When you differentiate a function, you find its rate of change. Conversely, when you integrate, you are looking for the original function that led to that rate of change. 

2.0Indefinite Integration formulas

• $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C;n\ne -1$
• $\int \frac{dx}{ax+b}=\frac{1}{a}\ln |ax+b|+C$
• $\int e^{ax+b}dx=\frac{1}{a}{e}^{ax+b}+C$
• $\int a^{px+q}dx=\frac{1}{p}\frac{a^{px+q}}{\ln a}+C,(a>0)$
• $\int \sin (ax+b)dx=-\frac{1}{a}\cos (ax+b)+C$
• $\int \cos (ax+b)dx=\frac{1}{a}\sin (ax+b)+C$
• $\int \tan (ax+b)dx=\frac{1}{a}\ln |\sec (ax+b)|+C$
• $\int \cot (ax+b)dx=\frac{1}{a}\ln |\sin (ax+b)|+C$
• $\int {\sec }^2(ax+b)dx=\frac{1}{a}\tan (ax+b)+C$
• $\int {\mathrm{cosec}}^2(ax+b)dx=-\frac{1}{a}\cot (ax+b)+C$
• $\int \mathrm{cosec}(ax+b)\cdot \cot (ax+b)dx=-\frac{1}{a}\mathrm{cosec}(ax+b)+C$
• $\int \sec (ax+b)\cdot \tan (ax+b)dx=\frac{1}{2}\sec (ax+b)+c$
• $\int \sec xdx=\ln |\sec x+\tan x|+C=\ln \left|\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\right|+C$
• $\int \mathrm{cosec}xdx=\ln |\mathrm{cosec}x-\cot x|+C=\ln \left|\tan \frac{x}{2}\right|+C=-\ln |\mathrm{cosec}x+\cot x|+C$
• $\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+C$
• $\int \frac{dx}{a^2+x^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C$
• $\int \frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}{\sec }^{-1}\frac{x}{a}+C$
• $\int \frac{dx}{\sqrt{x^2+a^2}}=\ln \left[x+\sqrt{x^2+a^2}\right]+C$
• $\int \frac{dx}{\sqrt{x^2-a^2}}=\ln \left[x+\sqrt{x^2-a^2}\right]+C$
• $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+C$
• $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln \left|\frac{x-a}{x+a}\right|+C$
• $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$
• $\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln \left(x+\sqrt{x^2+a^2}\right)+C$
• $\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln \left(x+\sqrt{x^2-a^2}\right)+C$
• $\int e^{ax}\cdot \sin bxdx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)+C=\frac{e^{ax}}{\sqrt{a^2+b^2}}\sin \left(bx-{\tan }^{-1}\frac{b}{a}\right)+C$
• $\int e^{ax}\cdot \cos bxdx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+C=\frac{e^{ax}}{\sqrt{a^2+b^2}}\cos \left(bx-{\tan }^{-1}\frac{b}{a}\right)+C$

3.0Indefinite Integration Methods

• Integration by Substitution
• Integration using Partial Fraction
• Integration By parts

Integration by Substitution: 

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 $\phi (x)=\mathrm{t}$  and ${\phi }^{\prime }(\mathrm{x})\mathrm{dx}=\mathrm{dt}$ .

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

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 & 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 as the function which comes first in the word ILATE, where; I-Inverse function, L-Logarithmic function, A-Algebraic function, T-Trigonometric function & E-Exponential function. 

Let 

$\begin{array}{rl}& \text{ I }=\int f(x)\cdot g(x)dx\\ & =f(x)\cdot \int g(x)dx-\int \left(f^{\prime }(x)\right)\left(\int g(x)dx\right)dx\end{array}$

= 1^st function × integral of 2^nd – $\int \text{ (diff. of }{1}^{\text{st }})\times \left(\text{ integral of }{2}^{\text{nd }}\right)dx$

3. $\int e^x\left(f(x)+f^{\prime }(x)\right)dx=e^{x}f(x)+c$
4. $\int e^{ax}\cdot \sin bxdx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)+c$
5. $\int e^{ax}\cdot \cos bxdx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+c$

Integration of Trigonometric Functions:

$\int {\sin }^{m}x{\cos }^{n}xdx$

Case-I: When m & n ∈ natural numbers.

• If one of them is odd, then substitute for the term of even power.
• If both are odd, substitute either of the term.
• If both are even, use trigonometric identities to convert integrand into cosines of multiple angles.

Case-II: m + n is a negative even integer.

• In this case the best substitution is tanx = t.

Integral of the form: 

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

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

In this case $\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2},\mathrm{x}=2{\tan }^{-1}\mathrm{t};dx=\frac{2dt}{1+t^2}$

Integral of the form:

$\int \frac{a\cos x+b\sin x+c}{p\cos x+q\sin x+r}dx$

Express Numerator $\left(N^r\right)=\ell \left(D^r\right)+m\left(D^r\right)+n$ & proceed.

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 px + q = l (differential coefficient of denominator) + m.

Integrals of the form:

$\int \frac{x^2+1}{x^4+K{x}^2+1}dx\text{ OR }\int \frac{x^2-1}{x^4+K{x}^2+1}dx$ where K is any constant.

Divide $N^r\&D^r$ by $x^2$ & proceed.

Note: Sometimes it is useful to write the integral as a sum of two related integrals, which can be evaluated by making suitable substitutions e.g.

• $\int \frac{2x^2}{x^4+1}dx=\int \frac{x^2+1}{x^4+1}dx+\int \frac{x^2-1}{x^4+1}dx$
• $\int \frac{2}{x^4+1}dx=\int \frac{x^2+1}{x^4+1}dx-\int \frac{x^2-1}{x^4+1}dx$

These integrals can be called as Algebraic Twins.

Integration of Irrational functions:

• $\int \frac{dx}{(ax+b)\sqrt{px+q}}$ and $\int \frac{dx}{\left(a{x}^2+bx+c\right)\sqrt{px+q}}\text{; put }px+q=t^2$
• $\int \frac{dx}{(ax+b)\sqrt{p{x}^2+qx+r}},\text{ put }ax+b=\frac{1}{t}\text{; }{\int }_1\frac{dx}{\left(a{x}^2+b\right)\sqrt{p{x}^2+q}},putx=\frac{1}{t}$

Manipulating integrands:

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

4.0What is the Difference Between Definite and Indefinite Integration?

Certainly! Here’s a tabular comparison between definite and indefinite integration:

5.0Indefinite Integrals Examples

Example 1: $\int (x+5{)}^{2}dx$

Solution:

M-1: $=\int \left(x^2+25+10x\right)dx=\frac{x^3}{3}+\frac{10x^2}{2}+25x+C$

M-2: $=\frac{(x+5{)}^3}{3}+C=\frac{x^3+15x^2+75x+125}{3}+C$

Example 2: Evaluate $\int \frac{dx}{8x+3}$

Solution: $\frac{1}{8}\ln |8x+3|+C$

Example 3: $\int {\cos }^{3}xdx$

Solution: $\int \frac{3\cos x+\cos 3x}{4}dx$ [cos3x = 4 cos³x – 3cosx]

$\Rightarrow \int \frac{3\cos x}{4}dx+\int \frac{\cos 3x}{4}dx$

$\Rightarrow \frac{3\sin x}{4}+\frac{\sin 3x}{12}+C$

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

Solution:

Put, tan^–1x = t

$\frac{1}{1+x^2}dx=dt$

$\Rightarrow \int \sin tdt$ = –cos t + C = –cos(tan^–1x) + C 

Example 5: $\int \frac{\sqrt{\tan x}}{\sin 2x}dx$

Solution:

$\int \frac{\sqrt{\tan x}}{2\tan x}\left(1+{\tan }^{2}x\right)dx\Rightarrow \int \frac{\sqrt{\tan x}}{2\tan x}{\sec }^{2}xdx$

Put tan x=t

sec²x dx = dt

$\frac{1}{2}\int (t{)}^{-1/2}dt=\frac{t^{1/2}\times 2}{2\times 1}+C$

$=\sqrt{t}+C$

$=\sqrt{\tan x}+C$

Example 6: $\int \frac{\sin xdx}{{\cos }^{2}x-36}$

Solution:

put cosx = t ⇒ –sinx dx = dt 

$\int \frac{-dt}{t^2-(6{)}^2}=-\frac{1}{2.6}\ln \left|\frac{t-6}{t+6}\right|+C$

$=-\frac{1}{12}\ln \left|\frac{\cos x-6}{\cos x+6}\right|+C$

Example 7: $\int e^x(\sin x+\cos x)dx$

Solution:

$\begin{array}{c}\int e^x(\sin x+\cos x)dx\end{array}\Rightarrow {\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x}+\mathrm{C}$

Example 8: Evaluate $I=\int e^x\cdot \sin xdx$

Solution:

Method-I:   

$I=\sin x\cdot e^x-\int \cos x\cdot e^{x}dx$

$\Rightarrow \mathrm{I}=\sin x{e}^x-\left[\cos x{e}^x-\int (-\sin x)e^{x}dx\right]$

$\Rightarrow I=\sin x{\mathrm{e}}^{\mathrm{x}}-\cos \mathrm{x}{\mathrm{e}}^{\mathrm{x}}-\mathrm{I}$

$\Rightarrow 2I=\sin x{e}^x-\cos x{e}^x$

$\Rightarrow I=\frac{1}{2}(\sin x-\cos x)e^x+\mathrm{c}$

Method-II : Assume $\int e^{ax}\cos bxdx=e^{ax}(A\cos bx+B\sin bx)$ and then differentiate both sides

Example 9: Evaluate $\int {\sin }^{2}x{\cos }^{4}xdx$

Solution:

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

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

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

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

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

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

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

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

Example 10: Evaluate: $\int \frac{2+3\cos \theta }{\sin \theta +2\cos \theta +3}d\theta$

Solution:

Write the Numerator = $\ell (denominator)$ + m(d.c. of denominator) + n 

⇒ 2 + 3 cosθ = l(sinθ + 2cosθ + 3) +m (cosθ – 2sinθ) +n.

Comparing the coefficients of sinθ, cosθ and constant terms, 

we get 3 l+n=2,  2l + m = 3, l – 2m = 0  

⇒  l = 6/5, m = 3/5 and n = –8/5

Hence,   $\mathrm{I}=\int \frac{6}{5}d\theta +\frac{3}{5}\int \frac{\cos \theta -2\sin \theta }{\sin \theta +2\cos \theta +3}d\theta -\frac{8}{5}\int \frac{d\theta }{\sin \theta +2\cos \theta +3}$

$=\frac{6}{5}\theta +\frac{3}{5}\ln |\sin \theta +2\cos \theta +3|-\frac{8}{5}{I}_3$ where 

${\mathbf{t}}_3=\int \frac{d\theta }{\sin \theta +2\cos \theta +3}$

In I₃, put $tan\frac{\theta }{2}=t\Rightarrow {\sec }^2\frac{\theta }{2}d\theta =2dt$

${\mathrm{I}}_3=2\int \frac{dt}{t^2+2t+5}=2\int \frac{dt}{(t+1{)}^2+2^2}=2.\frac{1}{2}{\tan }^{-1}\left(\frac{t+1}{2}\right)={\tan }^{-1}\left(\frac{\tan \theta /2+1}{2}\right)$

Hence 

$I=\frac{6\theta }{5}+\frac{3}{5}\ln |\sin \theta +2\cos \theta +3|-\frac{8}{5}{\tan }^{-1}\left(\frac{\tan \theta /2+1}{2}\right)+C$

6.0Practice Questions on Indefinite Integral

1. Solve $\int \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx$
2. Solve $\int \frac{\sin 2x+\sin 5x-\sin 3x}{\cos x+1-2{\sin }^{2}2x}dx$
3. Solve $\int \frac{d\varphi }{{\sin }^2\varphi {\cos }^2\varphi }$
4. Solve $\int \frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}dx$
5. Solve $\int \left(3\sin x{\cos }^{2}x-{\sin }^{3}x\right)dx$
6. Integrate with respect to $x:\frac{e^x+1}{e^x+x}$
7. Solve $\int \frac{d\left(\frac{x}{3}\right)}{\sqrt{1-{\left(\frac{x}{3}\right)}^2}}$
8. Solve $\int \frac{dx}{x^4\sqrt{x^2-3}}$
9. What is the power rule in indefinite integration?

Answers:  

1. $2[sinx+xcos\alpha ]+c$

2. –2(cosx) + c

3. –2cot2x + c

4. $\frac{1}{2}\sin 2x+c$

5. $-\frac{1}{3}\cos 3x+c$
6. ln (e^x + x) + c
7. ${\sin }^{-1}\left(\frac{x}{3}\right)+c$
8. $\frac{\mathrm{cosec}\left({\tan }^{-1}\frac{x}{\sqrt{3}}\right)}{9}-\frac{{\mathrm{cosec}}^3\left({\tan }^{-1}\frac{x}{\sqrt{3}}\right)}{27}+\mathrm{c}$
9. The power rule states that for any real number n ≠ –1, the indefinite integral of xⁿ is given by: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$