Integral of Particular Functions: Examples

Integration of indefinite integral is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of integration have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

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In this article, we will cover the concept of Integration of indefinite integral. This concept falls under the broader category of Calculus, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of two questions have been asked on this concept, including one in 2021, and one in 2023.

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.