Integral of Particular Functions: Examples

Integrals of Particular Functions

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 $y$ with respect to another quantity $x$ is called the derivative or differential coefficient of $y$ with respect to $x$. Geometrically, differentiation represents the slope of the tangent to a function’s graph, while integration determines the area under the curve of that function.

Basic Integration Formulas for Particular Functions

This section covers the most commonly used standard integration formulas that form the foundation for solving complex calculus problems. These basic integrals, such as those involving $\frac{1}{x^2 + a^2}$, $\frac{1}{x^2 - a^2}$, and $\sqrt{a^2 - x^2}$, help simplify advanced integrations through substitution or reduction techniques.

1. Integral of $\frac{1}{x^2 + a^2}$

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

Proof:

Put $x = a \tan \theta$, then $dx = a \sec^2 \theta , d\theta$

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

2. Integral of $\frac{1}{x^2 - a^2}$

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

Proof:

$\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \int \left(\frac{1}{x - a} - \frac{1}{x + a}\right) dx = \frac{1}{2a} [\log|x - a| - \log|x + a|] + C = \frac{1}{2a} \log \left|\frac{x - a}{x + a}\right| + C$

3. Integral of $\frac{1}{a^2 - x^2}$

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

4. Integral of $\frac{1}{\sqrt{a^2 - x^2}}$

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

Proof:

Let $x = a \sin \theta$, then $dx = a \cos \theta , d\theta$

$\int \frac{dx}{\sqrt{a^2 - x^2}} = \int \frac{a \cos \theta , d\theta}{a \cos \theta} = \int d\theta = \theta + C = \sin^{-1}\left(\frac{x}{a}\right) + C$

5. Integral of $\frac{1}{\sqrt{a^2 + x^2}}$

$\int \frac{dx}{\sqrt{a^2 + x^2}} = \log \left|x + \sqrt{a^2 + x^2}\right| + C$

Proof:

Let $x = a \tan \theta$, then $dx = a \sec^2 \theta , d\theta$

$\int \frac{dx}{\sqrt{a^2 + x^2}} = \int \frac{a \sec^2 \theta , d\theta}{a \sec \theta} = \int \sec \theta , d\theta = \log|\sec \theta + \tan \theta| + C$

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

6. Integral of $\frac{1}{\sqrt{x^2 - a^2}}$

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

Proof:

Let $x = a \sec \theta$, then $dx = a \sec \theta \tan \theta , d\theta$

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

$= \log|x + \sqrt{x^2 - a^2}| + C$

Integrals Involving Square Roots

Integrals containing square root expressions such as $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$ are common in problems related to geometry, physics, and engineering.

1. Integral of $\sqrt{a^2 - x^2}$

The standard result is:
$\int \sqrt{a^2 - x^2} , dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$

Proof:
Let $I = \int \sqrt{a^2 - x^2} , dx$

Substitute $x = a \sin \theta \Rightarrow dx = a \cos \theta , d\theta$

Then,
$I = \int \sqrt{a^2 - a^2 \sin^2 \theta} \cdot a \cos \theta , d\theta$
$= a^2 \int \cos^2 \theta , d\theta$

Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,
$I = \frac{a^2}{2} \int (1 + \cos 2\theta) , d\theta$

$= \frac{a^2}{2} \left(\theta + \frac{1}{2} \sin 2\theta\right) + C$

Now, substitute back $\sin \theta = \frac{x}{a}$ and $\sin 2\theta = 2 \sin \theta \cos \theta = \frac{2x\sqrt{a^2 - x^2}}{a^2}$

So,
$I = \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + \frac{x}{2} \sqrt{a^2 - x^2} + C$

Hence,
$\boxed{\int \sqrt{a^2 - x^2} , dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C}$

2. Integral of $\sqrt{a^2 + x^2}$

The standard result is:
$\int \sqrt{a^2 + x^2} , dx = \frac{x}{2} \sqrt{a^2 + x^2} + \frac{a^2}{2} \log \left|x + \sqrt{a^2 + x^2}\right| + C$

Proof:
Let $I = \int \sqrt{a^2 + x^2} , dx$

We use the substitution $x = a \tan \theta \Rightarrow dx = a \sec^2 \theta , d\theta$

Then,
$I = \int \sqrt{a^2 + a^2 \tan^2 \theta} \cdot a \sec^2 \theta , d\theta$
$= a^2 \int \sec^3 \theta , d\theta$

We know that $\int \sec^3 \theta , d\theta = \frac{1}{2}(\sec \theta \tan \theta + \log |\sec \theta + \tan \theta|) + C$

So,
$I = \frac{a^2}{2} (\sec \theta \tan \theta + \log |\sec \theta + \tan \theta|) + C$

Now, convert back to $x$:
$\tan \theta = \frac{x}{a}$ and $\sec \theta = \frac{\sqrt{a^2 + x^2}}{a}$

Thus,
$\sec \theta \tan \theta = \frac{x \sqrt{a^2 + x^2}}{a^2}$

Substituting,
$I = \frac{x}{2} \sqrt{a^2 + x^2} + \frac{a^2}{2} \log \left|x + \sqrt{a^2 + x^2}\right| + C$

Hence,
$\boxed{\int \sqrt{a^2 + x^2} , dx = \frac{x}{2} \sqrt{a^2 + x^2} + \frac{a^2}{2} \log \left|x + \sqrt{a^2 + x^2}\right| + C}$

3. Integral of $\sqrt{x^2 - a^2}$

The standard result is:
$\int \sqrt{x^2 - a^2} , dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log \left|x + \sqrt{x^2 - a^2}\right| + C$

Proof:
Let $I = \int \sqrt{x^2 - a^2} , dx$

Substitute $x = a \sec \theta \Rightarrow dx = a \sec \theta \tan \theta , d\theta$

Then,
$I = \int \sqrt{a^2 \sec^2 \theta - a^2} \cdot a \sec \theta \tan \theta , d\theta$
$= a^2 \int \tan^2 \theta \sec \theta , d\theta$

Using $\tan^2 \theta = \sec^2 \theta - 1$,
$I = a^2 \int (\sec^3 \theta - \sec \theta) , d\theta$

Integrate each term:
$\int \sec^3 \theta , d\theta = \frac{1}{2}(\sec \theta \tan \theta + \log |\sec \theta + \tan \theta|)$
and $\int \sec \theta , d\theta = \log |\sec \theta + \tan \theta|$

So,
$I = \frac{a^2}{2} (\sec \theta \tan \theta - \log |\sec \theta + \tan \theta|) + C$

Now, express in terms of $x$:
$\sec \theta = \frac{x}{a}$, $\tan \theta = \frac{\sqrt{x^2 - a^2}}{a}$

Then,
$\sec \theta \tan \theta = \frac{x \sqrt{x^2 - a^2}}{a^2}$

Substitute back:
$I = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log \left|x + \sqrt{x^2 - a^2}\right| + C$

Hence,
$\boxed{\int \sqrt{x^2 - a^2} , dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log \left|x + \sqrt{x^2 - a^2}\right| + C}$

Standard Trigonometric Substitutions

This section explains the standard trigonometric substitutions used to simplify integrals involving quadratic expressions like $a^2 - x^2$, $a^2 + x^2$, and $x^2 - a^2$. By converting algebraic forms into trigonometric ones, these substitutions make integration more straightforward and are essential for solving complex calculus problems efficiently.

These substitutions are crucial for simplifying irrational expressions and converting them into basic trigonometric integrals.

Working Rules for Substitution Method

1. For $\sqrt{a^2 - x^2}$, put $x = a \sin \theta$ or $x = a \cos \theta$

2. For $\sqrt{x^2 - a^2}$, put $x = a \sec \theta$ or $x = a \csc \theta$

3. For $\sqrt{a^2 + x^2}$, put $x = a \tan \theta$ or $x = a \cot \theta$

4. For $\sqrt{\frac{a - x}{a + x}}$, put $x = a \cos 2\theta$

5. Simplify and integrate using standard trigonometric identities

Solved Examples based on Integration of Particular Functions

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

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.

List of Topics Related to Integral of Particular Functions

Here, you’ll find a list of key topics associated with the integrals of particular functions. This section provides a structured overview to help you navigate through different types of integrals, their substitutions, and applications in problem-solving.

Application of Integrals

Integration as an inverse process of differentiation

Indefinite Integrals

Integration by Parts

Application of Inequality in Definite Integration

NCERT Resources

We have provided below the important NCERT resources including the NCERT Maths notes, exemplar solutions and exercises solutions for the Chapter 7 - Integrals to help you gain more understanding towards the chapter.

NCERT Class 12 Maths Notes for Chapter 7 - Integrals

NCERT Class 12 Maths Solutions for Chapter 7 - Integrals

NCERT Class 12 Maths Exemplar Solutions for Chapter 7 - Integrals

Practice Questions based on the Integral of Particular Functions

In this section, we have provided the practice questions based on integrals of particular functions along with topics related to it.

Integrals Of Particular Function- Practice Question MCQ

We have provided below the practice questions related to different concepts of integration to improve your understanding: