Integral of Particular Functions: Examples

Integrals of Particular Function

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. The rate of change of a quantity y concerning another quantity x is called the derivative or differential coefficient of y concerning x. Geometrically, the Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point.

1. $\int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$

$\begin{aligned}
& \text { put } x=a \tan \theta, \text { then } d x=a \sec ^2 \theta d \theta \\
& \begin{aligned}
\therefore \int \frac{d x}{x^2+a^2} & =\int \frac{a \sec ^2 \theta}{a^2+a^2 \tan ^2 \theta} d \theta \\
& =\int \frac{a^2 \sec ^2 \theta}{a^2\left(1+\tan ^2 \theta\right)} d \theta \\
& =\frac{1}{a} \int d \theta=\frac{1}{a} \theta+C=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C
\end{aligned}
\end{aligned}$

2. $\int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$
we can rewrite above integral as

$\begin{aligned}
\int \frac{d x}{x^2-a^2} & =\frac{1}{2 a} \int\left(\frac{1}{x-a}-\frac{1}{x+a}\right) d x \\
& =\frac{1}{2 a}(\log |x-a|-\log |x+a|)+c \\
& =\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C
\end{aligned}$

3. $\int \frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C$

4. $\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1}\left(\frac{x}{a}\right)+C$

5. $\int \frac{d x}{\sqrt{\mathrm{a}^2+\mathrm{x}^2}}=\log \left|x+\sqrt{\mathrm{x}^2+\mathrm{a}^2}\right|+C$

6. $\int \frac{d x}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+C$

7. $\int \sqrt{a^2-x^2} d x=\frac{1}{2} x \sqrt{a^2-x^2}+\frac{1}{2} a^2 \sin ^{-1}\left(\frac{x}{a}\right)+C$

8. $\int \sqrt{a^2+x^2} d x=\frac{1}{2} x \sqrt{a^2+x^2}+\frac{1}{2} a^2 \log \left|x+\sqrt{a^2+x^2}\right|+C$

9. $\int \sqrt{\mathrm{x}^2-\mathrm{a}^2} \mathrm{dx}=\frac{1}{2} \mathrm{x} \sqrt{\mathrm{x}^2-\mathrm{a}^2}-\frac{1}{2} \mathrm{a}^2 \log \left|\mathrm{x}+\sqrt{\mathrm{x}^2-\mathrm{a}^2}\right|+C$

Following are some important substitutions useful in evaluating integrals.

$\begin{array}{c|| c } \mathbf { Expression } & {\mathbf { Substitution }} \\\\ \hline \\a^{2}+x^{2} & {x=a \tan \theta} {\text { or }} {x=a \cot \theta} \\ \\\hline \\a^{2}-x^{2} & {x=a \sin \theta} {\text { or } x=a \cos \theta}\\ \\ \hline \\x^{2}-a^{2} & {x=a \sec \theta} {\text { or } x=a \csc \theta} \\\\ \hline\\ \sqrt{\frac{a-x}{a+x}} {\text { or } \sqrt{\frac{a+x}{a-x}}} & {x=a \cos 2 \theta}\\ \\\hline\end{array}$

A special type of indefinite integration:

Working rule :
for (i) put $x=a \sin \theta$ or $a \cos \theta$
for (ii) Put $x=a \sec \theta$ or $a \operatorname{cosec} \theta$
for (iii) and (iv) Put $x=a \tan \theta$ or $a \cot \theta$
for (v) and (vi) Put $x=a \cos 2 \theta$
for (vii) and (viii) Put $x=a \cos ^2 \theta+b \sin ^2 \theta$
Integrals of the form :
(i) $\int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$
(ii) $\int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$
(iii) $\int \frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C$
(iv) $\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1}\left(\frac{x}{a}\right)+C$
(v) $\int \frac{d x}{\sqrt{a^2+x^2}}=\log \left|x+\sqrt{x^2+a^2}\right|+C$
(vi) $\int \frac{d x}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+C$

Proofs of all these formulas are as follows:

Integrals of Some Particular Functions

In this section, we mention below some important formulas of integrals and apply them for integrating many other related standard integrals:

(1) $\int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+\mathrm{C}$
proofs of all these formulas area as follows:
We have $\frac{1}{x^2-a^2}-\frac{1}{(x-a)(x+a)}$

$\equiv \frac{1}{2 a}\left[\frac{(x+a)-(x-a)}{(x-a)(x+a)}\right]-\frac{1}{2 a}\left[\frac{1}{x-a}-\frac{1}{x+a}\right]$

Therefore, $\int \frac{d x}{x^2-a^2}=\frac{1}{2 a}\left[\int \frac{d x}{x-a}-\int \frac{d x}{x+a}\right]$

$\begin{aligned}
& \left.=\frac{1}{2 a}[\log |(x-a)|-\log \mid(x+a)]\right]+\mathrm{C} \\
& =\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+\mathrm{C}
\end{aligned}$

(2) In view of (1) above, we have

$\frac{1}{a^2-x^2}=\frac{1}{2 a}\left[\frac{(a+x)+(a-x)}{(a+x)(a-x)}\right]=\frac{1}{2 a}\left[\frac{1}{a-x}+\frac{1}{a+x}\right]$

Therefore, $\int \frac{d x}{a^2-x^2}=\frac{1}{2 a}\left[\int \frac{d x}{a-x}+\int \frac{d x}{a+x}\right]$

$\begin{aligned}
& -\frac{1}{2 a}[-\log |a-x|+\log |a+x|]+\mathrm{C} \\
& -\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+\mathrm{C}
\end{aligned}$

- Note The technique used in (1) will be explained in Section 7.5.

(3) Put $x=a \tan \theta$. Then $d x=a \sec ^2 \theta d \theta$.

Therefore, $\int \frac{d x}{x^2+a^2}-\int \frac{a \sec ^2 \theta d \theta}{a^2 \tan ^2 \theta+a^2}$

(4) Let $x=a \sec \theta$. Then $d x=a \sec \theta \tan \theta d \theta$.

$-\frac{1}{a} \int d \theta=\frac{1}{a} \theta+\mathrm{C}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+\mathrm{C}$

Therefore, $\quad \int \frac{d x}{\sqrt{x^2-a^2}}=\int \frac{a \sec \theta \tan \theta d \theta}{\sqrt{a^2 \sec ^2 \theta-a^2}}$
$-\int \sec \theta d \theta=\log |\sec \theta+\tan \theta|+C_1$

$=\log \left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right|+C_1$

$-\log \left|x+\sqrt{x^2-a^2}\right|-\log |a|+C_1$

$=\log \left|x+\sqrt{x^2-a^2}\right|+C, \text { where } C=C_1-\log |a|$

(5) Let $x=a \tan \theta$. Then $d x=a \sec ^2 \theta \mathrm{d} \theta$.

$\begin{aligned}
\int \frac{d x}{\sqrt{a^2-x^2}} & =\int \frac{a \cos \theta d \theta}{\sqrt{a^2-a^2 \sin ^2 \theta}} \\
& =\int d \theta-\theta+\mathrm{C}-\sin ^{-1} \frac{x}{a}+\mathrm{C}
\end{aligned}$
Therefore, $\quad \int \frac{d x}{\sqrt{x^2+a^2}}=\int \frac{a \sec ^2 \theta d \theta}{\sqrt{a^2 \tan ^2 \theta+a^2}}$
$=\int \sec \theta d \theta-\log |(\sec \theta+\tan \theta)|+C_1$
$\begin{aligned}
& =\log \left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}+1}\right|+C_1 \\
& =\log \left|x+\sqrt{x^2+a^2}\right|-\log |a|+C_1 \\
& =\log \left|x+\sqrt{x^2+a^2}\right|+C, \text { where } C-C_1-\log |a|
\end{aligned}$

Recommended Video Based on Integrals of Particular Function

Solved Examples

Example 1: Evaluate $\int \frac{d x}{\sqrt{(x-a)(b-x)}}$

1) $2 \sin ^{-1} \sqrt{\left(\frac{x-a}{b-a}\right)}+c$

2) $2 \cos ^{-1} \sqrt{\left(\frac{x-a}{b-a}\right)}+c$

3) $2 \tan ^{-1} \sqrt{\left(\frac{x-a}{b-a}\right)}+c$

4) $2 \tan ^{-1} \sqrt{\left(\frac{x-a}{b-x}\right)}+c$

Solution

Writing $x=a \cos ^2 \theta+b \sin ^2 \theta=a+(b-a) \sin ^2 \theta$, the given integral becomes

$\begin{aligned}
& I=\int \frac{2(b-a) \sin \theta \cos \theta d \theta}{\left\{\left(a \cos ^2 \theta+b \sin ^2 \theta-a\right)\left(a \cos ^2 \theta+b \sin ^2 \theta-b\right)\right\}^{1 / 2}} \\
& =\int \frac{2(b-a) \sin \theta \cos \theta d \theta}{(b-a) \sin \theta \cos \theta}=\left(\frac{b-a}{b-a}\right) \int 2 d \theta
\end{aligned}$

$=2 \theta+c=2 \sin ^{-1} \sqrt{\left(\frac{x-a}{b-a}\right)}+c$
Hence, the answer is the option 1.

Example 2: Evaluate $\int \ln (\sqrt{1-x}+\sqrt{1+x}) d x$

1) $x \ln (\sqrt{1-x}+\sqrt{1+x})+\frac{x}{2}+\frac{1}{2} \sin ^{-1} x+c$

2) $x \ln (\sqrt{1-x}+\sqrt{1+x})-\frac{x}{2}+\frac{1}{2} \sin ^{-1} x+c$

3) $x \ln (\sqrt{1-x}+\sqrt{1+x})-\frac{x}{2}-\frac{1}{2} \sin ^{-1} x+c$

4) $x \ln (\sqrt{1-x}+\sqrt{1+x})-\frac{x}{2}+\frac{1}{2} \sin ^{-1} x+c$

Solution

We can do this question using Integration by parts

If we take

$u=\ln (\sqrt{1-x}+\sqrt{1+x}) \text { as the first function and } \mathrm{v}=1 \text { as the second function then }$

$\begin{aligned}
& \ln (\sqrt{1-x}+\sqrt{1+x}) \int 1 d x-\int\left(\frac{d}{d x}(\ln (\sqrt{1-x}+\sqrt{1+x})) \int 1 d x\right) d x \\
& =x \ln (\sqrt{1-x}+\sqrt{1+x})-\int \frac{1}{(\sqrt{1-x}+\sqrt{1+x})}\left(-\frac{1}{2 \sqrt{1-x}}+\frac{1}{2 \sqrt{1+x}}\right) x d x=x \ln (\sqrt{1-x}+\sqrt{1+x})-\frac{1}{2} \int x \frac{\sqrt{1-x^2}}{x \sqrt{1-x^2}} d x \\
& =x \ln (\sqrt{1-x}+\sqrt{1+x})-\frac{1}{2} \int d x+\frac{1}{2} \int \frac{1}{\sqrt{1-x^2}} d x
\end{aligned}$
$=x \ln (\sqrt{1-x}+\sqrt{1+x})-\frac{x}{2}+\frac{1}{2} \sin ^{-1} x+c$
Hence, the answer is the option (2).

Example 3: $\int\left(\frac{2 a+x}{a+x}\right) \sqrt{\frac{a-x}{a+x}} d x=$
1) $\sqrt{a^2-x^2}-2 a \sqrt{\frac{a-x}{a+x}}+c$
2) $-\sqrt{a^2-x^2}-2 a \sqrt{\frac{a-x}{a+x}}+c$
3) $\frac{1}{a} \tan ^{-1} \frac{x}{a}+\ln \left|x+\sqrt{a^2-x^2}\right|+c$
4) $\frac{1}{2 a} \ln \left|\frac{a+x}{a-x}\right|+\sin ^{-1} \frac{x}{a}+c$

Solution
As we learnt in
Special types of indefinite integration:
Integrals of the form:

$f\left(\sqrt{\frac{a-x}{a+x}}\right)_{\text {(ii) }} f\left(\sqrt{\frac{a+x}{a-x}}\right)$

wherein

The working rule is :

for (i) and (ii) Put $x=a \cos ($

$\begin{aligned}
& \quad \theta=\cos ^{-1} \frac{x}{a}(-a<x<a) \\
& \text { Put } I=-a \int \frac{(2+\cos \theta)(1-\cos \theta)}{1+\cos \theta} d \theta \\
& \text { and } \\
& =-a \int\left\{(1-\cos \theta)+\frac{1-\cos \theta}{1+\cos \theta}\right\} d \theta=-a\left(2 \tan \frac{\theta}{2}-\sin \theta\right)+c \\
& =\sqrt{a^2-x^2}-2 a \sqrt{\frac{a-x}{a+x}}+c
\end{aligned}$

Hence, the answer is the option 1.

Example 4: $\int \frac{d x}{x \sqrt{1-x^3}}=$
1) $\frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+c$
2) $\frac{1}{3} \log \left|\frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}+1}\right|+c$
3) $\frac{1}{3} \log \left|\frac{1}{\sqrt{1-x^3}}\right|+c$
4) $\frac{1}{3} \log \left|1-x^3\right|+c$

Solution

As we learned,

$\int \frac{d x}{x \sqrt{1-x^3}}$

Put $1-x^3=t^2$

$\begin{aligned}
& -3 x^2 \mathrm{dx}=2 \text { tdt } \\
& =-\frac{2}{3} \int \frac{d t}{1-t^2}=\frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+c
\end{aligned}$

Hence, the answer is the option 1.

Example 5: if $\int \sqrt{\frac{\cos x-\cos ^3 x}{\left(1-\cos ^3 x\right)}} d x=f(x)+c$, then $f(x)$ is equal to
1) $\frac{2}{3} \sin ^{-1}\left(\cos ^{3 / 2} x\right)$
2) $\frac{3}{2} \sin ^{-1}\left(\cos ^{3 / 2} x\right)$
3) $\frac{2}{3} \cos ^{-1}\left(\cos ^{3 / 2} x\right)$
4) None of these

4) None of these

Solution

As we learned in

Integration of Rational and irrational functions -

Integration in the form of :

$\begin{aligned} & \frac{d x}{\sqrt{a^2-x^2}} \\ & I=\int \sqrt{\frac{\cos x-\cos ^3 x}{1-\cos ^3 x}} d x=\int \frac{\sqrt{\cos x} \sqrt{1-\cos ^2 x}}{\sqrt{1-\left(\cos ^{3 / 2} x\right)^2}} d x \\ & \int \frac{\sqrt{\cos x} \sin x}{\sqrt{1-\left(\cos ^{3 / 2} x\right)^2}} d x \\ & \text { If } \cos ^{\frac{3}{2}} x=p, \text { then }\left(-\frac{3}{2} \cos ^{\frac{1}{2}} x \sin x\right) d x=d p \\ & I=-\frac{2}{3} \int \frac{d p}{\sqrt{1-p^2}}=-\frac{2}{3} \sin ^{-1}\left(\cos ^{\frac{3}{2} x}\right)=\frac{2}{3} \cos ^{-1}\left(\cos ^{\frac{3}{2}} x\right)+c_1\end{aligned}$

Hence, the answer is the option 3.

Summary

An indefinite integral, also known as an antiderivative, is a function that reverses the process of differentiation. It's an important concept of the calculus. In physics, integration is used to calculate quantities such as work, energy, and centre of mass.