Integrals of Particular Functions

Integration is a method to sum functions on a large scale. Here, we discuss integrals of some particular functions commonly used in calculations. These integrals have significant real-life applications, such as finding areas between curves, volumes, average values of functions, kinetic energy, centers of mass, and work done, among others.

1.0Integrals of Particular Functions

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

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

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

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

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

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

2.0Proofs of Integral Functions

1. We have $\frac{1}{x^2-a^2}=\frac{1}{(x-a)(x+a)}$

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

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

$=\frac{1}{2a}[\log |(x-a)|-\log |(x+a)|]+C$

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

2. In view of (1) above, we have 

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

Therefore,

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

$=\frac{1}{2a}[-\log |a-x|+\log |a+x|]+C$

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

3. Put x = a tan θ. Then dx = a sec² θ dθ.

Therefore,

$\int \frac{dx}{x^2+a^2}=\int \frac{a^2{\sec }^2\theta d\theta }{a^2{\tan }^2\theta +a^2}$

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

4. Let x = a sec θ. Then dx = a sec θ tan θ dθ.

Therefore,

$\int \frac{dx}{\sqrt{x^2-{\mathrm{a}}^2}}=\int \frac{\mathrm{a}\sec \theta \tan \theta d\theta }{\sqrt{{\mathrm{a}}^2{\sec }^2\theta -{\mathrm{a}}^2}}$

$=\int \sec \theta d\theta =\log |\sec \theta +\tan \theta |+C_1$

$=\log \left|\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-1}\right|+{\mathrm{C}}_1$

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

$=\log \left|\mathrm{x}+\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}\right|+\mathrm{C}\text{, where }\mathrm{C}={\mathrm{C}}_1-\log |\mathrm{a}|$

5. Let x = a sinθ. Then dx = a cos θ dθ.

Therefore,

$\int \frac{dx}{\sqrt{a^2-x^2}}=\int \frac{a\cos \theta d\theta }{\sqrt{a^2-a^2{\sin }^2\theta }}$

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

6. Let x = a tan θ. Then dx = a sec² θ dθ.

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

$=\int \sec \theta d\theta =\log |(\sec \theta +\tan \theta )|+C_1$

$=\log \left|\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+1}\right|+{\mathrm{C}}_1$

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

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

Applying these standard formulae, we now obtain some more formulae which are useful from applications point of view and can be applied directly to evaluate other integrals.

3.0Some more Integrals

1. To find the integral $\int \frac{dx}{a{x}^2+bx+c}$ , we write

$a{x}^2+bx+c=a\left[x^2+\frac{b}{a}x+\frac{c}{a}\right]=a\left[{\left(x+\frac{b}{2a}\right)}^2+\left(\frac{c}{a}-\frac{b^2}{4a^2}\right)\right]$

Now, put $\mathrm{x}+\frac{\mathrm{b}}{2\mathrm{a}}=\mathrm{t}$ so that dx = dt and writing $\frac{c}{a}-\frac{b^2}{4a^2}=\pm k^2$ . We find the integral reduced to the form $\frac{1}{a}\int \frac{dt}{t^2\pm k^2}$ depending upon the sign of $\left(\frac{c}{a}-\frac{b^2}{4a^2}\right)$ and hence can be evaluated.

2. To find the integral of the type $\int \frac{dx}{\sqrt{a{x}^2+bx+c}}$ , proceeding as in (1), we obtain the integral using the standard formulae.
3. To find the integral of the type $\int \frac{px+q}{a{x}^2+bx+c}dx$, where given p, q, a, b, c are constants. Now, we have to find real numbers, A, B such that:

$\mathrm{px}+\mathrm{q}=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\right)+\mathrm{B}=\mathrm{A}(2\mathrm{ax}+\mathrm{b})+\mathrm{B}$

To determine A and B, we equate the coefficients of x and the constant term from both sides. By solving these equations, we obtain A and B, thus reducing the integral to a known form.

4. For the evaluation of the integral of the type $\int \frac{(px+q)dx}{\sqrt{a{x}^2+bx+c}}$, we proceed as in (3) and reduce the integral into known standard forms.

4.0Difference Between a Particular Solution and a Particular Integral

5.0Solved Examples of Integrals of Particular Functions

Example 1: $\int \frac{\sin xdx}{{\cos }^{2}x-36}$

Solution: put cos x = t ⇒ –sin x 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 2: $\int \frac{dx}{3x^2+6x+15}$

Solution: ⇒ $\frac{1}{3}\int \frac{dx}{x^2+2x+5}$

⇒ $\frac{1}{3}\int \frac{dx}{(x+1{)}^2+4}=\frac{1}{3}\int \frac{dx}{(x+1{)}^2+(2{)}^2}$

$=\frac{1}{6}{\tan }^{-1}\left(\frac{x+1}{2}\right)+C$

Example 3: $\int \frac{dx}{\sqrt{x^2-5x}}$

Solution: $\int \frac{dx}{\sqrt{{\left(x-\frac{5}{2}\right)}^2-{\left(\frac{5}{2}\right)}^2}}$ 

$=\ln \left[\left(x-\frac{5}{2}\right)+\sqrt{{\left(x-\frac{5}{2}\right)}^2-{\left(\frac{5}{2}\right)}^2}\right]+C$

Example 4: $\int \frac{2x}{\sqrt{x^4+2x^2+4}}dx$

Solution: put x² = t   ⇒  2x dx = t 

$\int \frac{dt}{\sqrt{t^2+2t+4}}=\int \frac{dt}{\sqrt{(t+1{)}^2+(\sqrt{3}{)}^2}}$

$=\ln \left|(t+1)+\sqrt{(t+1{)}^2+(\sqrt{3}{)}^2}\right|+C$

t = x²

Example 5: Evaluate $\int \frac{dx}{\sqrt{(x-a)(b-x)}}$

Solution: Put x = a cos²θ + b sin²θ, the given integral becomes

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

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

Example 6: Evaluate $\int \frac{dx}{2x^2+x-1}$

Solution:

$I=\int \frac{dx}{2x^2+x-1}=\frac{1}{2}\int \frac{dx}{x^2+\frac{x}{2}-\frac{1}{2}}=\frac{1}{2}\int \frac{dx}{x^2+\frac{x}{2}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}}$

$=\frac{1}{2}\int \frac{dx}{(x+1/4{)}^2-9/16}=\frac{1}{2}\int \frac{dx}{(x+1/4{)}^2-(3/4{)}^2}$

$=\frac{1}{2}\cdot \frac{1}{2(3/4)}\log \left|\frac{x+1/4-3/4}{x+1/4+3/4}\right|+C$

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

$=\frac{1}{3}\log \left|\frac{x-1/2}{x+1}\right|+C=\frac{1}{3}\log \left|\frac{2x-1}{2(x+1)}\right|+C$

Ans.

Example 7: Evaluate $\int \frac{3x+2}{4x^2+4x+5}dx$

Solution: Express 3x + 2 = ℓ (d.c. of 4x² + 4x + 5) + m

or, 3x + 2 = ℓ (8x + 4) + m

Comparing the coefficients, we get

8ℓ = 3 and 4ℓ + m = 2 ⇒ ℓ = 3/8 and m = 2 – 4ℓ = 1/2

⇒  $I=\frac{3}{8}\int \frac{8x+4}{4x^2+4x+5}dx+\frac{1}{2}\int \frac{dx}{4x^2+4x+5}$

$=\frac{3}{8}\log \left|4x^2+4x+5\right|+\frac{1}{8}\int \frac{dx}{x^2+x+\frac{5}{4}}$

$=\frac{3}{8}\log \left|4x^2+4x+5\right|+\frac{1}{8}{\tan }^{-1}\left(x+\frac{1}{2}\right)+C$

Ans.

6.0Practice Questions of Integrals of Particular Functions

a. $\int \frac{dx}{2x^2+9}$

b. $\int \frac{dx}{\sqrt{4-x^2}}$

c. $\int \frac{dx}{1+9x^2}$

d. $\int \frac{dx}{\sqrt{1-25x^2}}$

e. $\int \frac{dx}{25+4x^2}$

f. $\int \frac{x^2}{9+16x^6}dx$

g. $\int \frac{dx}{x\sqrt{x^2+4}}$

h. $\int \frac{dx}{\sqrt{4x-3-x^2}}$

i. $\int \frac{\cos \alpha d\alpha }{a^2+{\sin }^2\alpha }$

j. $\int \frac{xdx}{x^4+1}$

Answers:

a. $\frac{1}{3\sqrt{2}}\arctan \frac{\sqrt{2}}{3}x+C$

b. $\mathrm{arc}\sin \frac{x}{2}+C$

c. $\frac{1}{3}\mathrm{arc}\tan 3x+C$

d. $\frac{1}{5}\mathrm{arc}\sin 5x+C$

e. $\frac{1}{10}{\tan }^{-1}\frac{2x}{5}+C$ 

f. $\frac{1}{36}{\tan }^{-1}\left(\frac{4x^3}{3}\right)+C$

g. $\frac{1}{2}\ln \left[\frac{\sqrt{x^2+4}-2}{x}\right]+C$

h. $\mathrm{arc}\sin (x-2)+C$

i. $\frac{1}{a}\arctan \frac{\sin \alpha }{a}+C$

j. $\frac{1}{2}\arctan x^2+C$

7.0Sample Questions on Integrals of Particular Functions

Q. What is the integral of $\frac{1}{x^2-a^2}$ ?

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

Q. What is the integral of $\frac{1}{x^2+a^2}$ ?

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