# Indirect Substitution in Integral

## Introduction To Indirect Substitution

A function which takes the antiderivative of another function is an indefinite integral. It is denoted by the symbol of integration, a function and dx at the end. The indefinite integral is a simpler way to represent the antiderivative. The three methods to find the value of an indefinite integral are as follows:

• Integration by substitution [direct and indirect]
• Integration by partial fractions
• Integration by parts

### Indirect Substitutions

• Suppose the integrand of f (x) can be written as the product of the functions f (x) and h (x), where h (x) being the function of the integral f (x), then the integral f (x) can be put as t.
• If any function ∫ f (x) dx = g (x), then ∫ f (ax + b) = [1 / a] g (ax + b)
• The integral of the form ∫ [f’ (x) / f (x)] dx = log (f (x)) + c
• The integral ∫ [f’ (x) / √ f (x)] dx = 2 √[f (x)] + c
• ∫ [[f (x)]n f’ (x)] dx = {[f (x)]n+1 / (n+1) + C}, where n ≠ 1

If integrals of the form ∫sinpx dx, ∫cosqx dx [p less than or equal to 4], sinpx and cosqx are expressed in terms of sines and cosines of multiples of x using the identities of trigonometry.

Steps to evaluate integrals of the form sinpx and cosqxdx,

1] Ensure whether the exponents of sinx and cosx belong to the set of natural numbers.

2] The different cases for exponents are as follows:

a] Say the exponent of sinx is odd, then take cosx = t.

b] If the exponent of cosx is odd, then sinx = t is taken.

c] If both the exponents are odd, then either sinx or cosx is taken as variable “t”.

d] In the case of both the exponents being positive, they are expressed in terms of sines and cosines of x using the trigonometric identities.

### Derived Substitutions

It is easy to compute if one integral is written as the sum of two related integrals by making appropriate substitutions.

• Algebraic twins

• Trigonometric twins

## Indirect Substitution In Integral Problems

Example 1: Solve the integral $\int \frac{d({x^2}+1)}{^{\sqrt{x^2+2}}}$

Solution:

$I=\int \frac{d({x^2}+1)}{^{\sqrt{x^2+2}}}$

We know,

$d(x^2+1)=2x\ dx\\I=\int \frac{2xdx}{\sqrt{x^2+2}}\\x^2+2=t^2\\ 2x\ dx=2t\ dt\\ I=\int \frac{2t\ dt}{t}=2t+C I=2\sqrt{x^2+2}+C$

Example 2: ∫ secx log (sec x + tan x) dx

Solution:

Let log (sec x + tan x) = t

⇒ sec x dx = dt

Therefore, ∫ sec x log (sec x + tan x) = ∫ t dt

= t2 / 2 + c

= [log (sec x + tan x)]2 / 2 + c

Example 3: ∫ [sin 2x] / [sin4 x + cos4 x] dx

Solution:

∫ [sin 2x] / [sin4 x + cos4 x] dx =∫[2 sin x cos x] /[sin4 x + cos4 x] dx

=∫[2 tan x sec2x] / [1 + tan4x] dx

Put tan2x = t ⇒2 tan x sec2 x dx = dt, then it reduced to ⇒ f (x) = x3 / 3 + 5x + c.

Hint : By inspection,

d / dx {cot−1 (tan2x)} = − [1 (2 tan x . sec2x)] / [1 + tan4 x] = − sin 2x / sin4 x + cos4 x

⇒ d / dx {tan−1(tan2x)} = sin 2x / [sin4 x + cos4 x]

Example 4: ∫ ([x + 1] / √[1 + x2]) dx

Solution:

∫ ([x + 1] / √[1 + x2]) dx =∫(x / √[x2+1]) dx + ∫(1 / √[x2 + 1]) dx

Put x2 + 1 = t ⇒ 2x dx = dt, then it reduces to

[1 / 2] ∫dt / t1/2 + ∫1 / √[x2 + 1] dx = 1 / 2 * 2 * t1/2 + log (x + √[x2 + 1]) + c

= (x2 + 1)½ + log (x + √[x2 + 1]) + c

Example 5: ∫ tan (3x − 5) sec(3x − 5) dx

Solution:

Put t = 3x − 5 ⇒ dt = 3 dx, therefore

∫tan (3x − 5) sec (3x − 5) dx = 1 / 3 ∫[tant * sect] dt

= sect / 3 + c

= sec (3x − 5) / 3 + c

Example 6: Evaluate the integral $\int \frac{5}{1+x^4}dx$

Solution:

$=5\cdot \int \frac{1}{1+x^4}dx\\\mathrm{Take\:the\:partial\:fraction\:of}\:\frac{1}{1+x^4}\\=5\cdot \int \frac{\sqrt{2}-x}{2\sqrt{2}\left(x^2-\sqrt{2}x+1\right)}+\frac{x+\sqrt{2}}{2\sqrt{2}\left(x^2+\sqrt{2}x+1\right)}dx\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\=5\left(\int \frac{\sqrt{2}-x}{2\sqrt{2}\left(x^2-\sqrt{2}x+1\right)}dx+\int \frac{x+\sqrt{2}}{2\sqrt{2}\left(x^2+\sqrt{2}x+1\right)}dx\right)\\\int \frac{\sqrt{2}-x}{2\sqrt{2}\left(x^2-\sqrt{2}x+1\right)}dx\\=\frac{1}{2\sqrt{2}}\cdot \int \frac{\sqrt{2}-x}{x^2-\sqrt{2}x+1}dx\\x^2-\sqrt{2}x+1\\\mathrm{Write}\:x^2-\sqrt{2}x+1\:\mathrm{in\:the\:form:\:\:}x^2+2cx+c^2\\2c=-\sqrt{2}\quad :\quad c=-\frac{\sqrt{2}}{2}\\=x^2-\sqrt{2}x+1+\left(-\frac{\sqrt{2}}{2}\right)^2-\left(-\frac{\sqrt{2}}{2}\right)^2\\=\left(x-\frac{\sqrt{2}}{2}\right)^2+1-\left(-\frac{\sqrt{2}}{2}\right)^2\\=\left(x-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}\\=\frac{1}{2\sqrt{2}}\cdot \int \frac{\sqrt{2}-x}{\left(x-\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}dx\\\mathrm{Apply\:u-substitution:}\:u=x-\frac{\sqrt{2}}{2}\\=\frac{1}{2\sqrt{2}}\cdot \int \frac{\sqrt{2}\left(-\sqrt{2}u+1\right)}{2u^2+1}du\\=\frac{1}{2\sqrt{2}}\sqrt{2}\cdot \int \frac{-\sqrt{2}u+1}{2u^2+1}du\\=\frac{1}{2\sqrt{2}}\sqrt{2}\left(-\int \frac{\sqrt{2}u}{2u^2+1}du+\int \frac{1}{2u^2+1}du\right)\\=\frac{1}{2\sqrt{2}}\sqrt{2}\left(-\frac{1}{2\sqrt{2}}\ln \left|2u^2+1\right|+\frac{1}{\sqrt{2}}\arctan \left(\sqrt{2}u\right)\right)\\\mathrm{Substitute\:back}\:u=x-\frac{\sqrt{2}}{2}\\=\frac{1}{2\sqrt{2}}\sqrt{2}\left(-\frac{1}{2\sqrt{2}}\ln \left|2\left(x-\frac{\sqrt{2}}{2}\right)^2+1\right|+\frac{1}{\sqrt{2}}\arctan \left(\sqrt{2}\left(x-\frac{\sqrt{2}}{2}\right)\right)\right)\\=\frac{1}{4\sqrt{2}}\left(-\ln \left|2x^2-2\sqrt{2}x+2\right|+2\arctan \left(\sqrt{2}x-1\right)\right)\\=5\left(\frac{1}{4\sqrt{2}}\left(-\ln \left|2x^2-2\sqrt{2}x+2\right|+2\arctan \left(\sqrt{2}x-1\right)\right)+\frac{1}{4\sqrt{2}}\left(\ln \left|2x^2+2\sqrt{2}x+2\right|+2\arctan \left(\sqrt{2}x+1\right)\right)\right)\\=5\left(\frac{1}{4\sqrt{2}}\left(-\ln \left|2x^2-2\sqrt{2}x+2\right|+2\arctan \left(\sqrt{2}x-1\right)\right)+\frac{1}{4\sqrt{2}}\left(\ln \left|2x^2+2\sqrt{2}x+2\right|+2\arctan \left(\sqrt{2}x+1\right)\right)\right)\\$

Example 7: Solve the integral $\int \sqrt{tanx}dx$

Solution:

$\int \sqrt{tanx}dx\\\mathrm{Apply\:Trig\:Substitution:}\:x=\arctan \left(u\right)\\=\int \frac{\sqrt{u}}{1+u^2}du\\=\int \frac{\sqrt{2}\sqrt{v}}{4+v^2}dv\\=\sqrt{2}\cdot \int \frac{\sqrt{v}}{4+v^2}dv\\\mathrm{Apply\:u-substitution:}\:w=\sqrt{v}\\=\sqrt{2}\cdot \int \frac{2w^2}{4+w^4}dw\\=\sqrt{2}\cdot \:2\cdot \int \frac{w^2}{4+w^4}dw\\\mathrm{Take\:the\:partial\:fraction\:of}\:\frac{w^2}{4+w^4}\\=\sqrt{2}\cdot \:2\cdot \int \frac{w}{4\left(w^2-2w+2\right)}-\frac{w}{4\left(w^2+2w+2\right)}dw\\=\sqrt{2}\cdot \:2\left(\int \frac{w}{4\left(w^2-2w+2\right)}dw-\int \frac{w}{4\left(w^2+2w+2\right)}dw\right)\\\int \frac{w}{4\left(w^2-2w+2\right)}dw=\frac{1}{8}\left(\ln \left|w^2-2w+2\right|+2\arctan \left(w-1\right)\right)\\\int \frac{w}{4\left(w^2+2w+2\right)}dw=\frac{1}{8}\left(\ln \left|w^2+2w+2\right|-2\arctan \left(w+1\right)\right)\\=\sqrt{2}\cdot \:2\left(\frac{1}{8}\left(\ln \left|w^2-2w+2\right|+2\arctan \left(w-1\right)\right)-\frac{1}{8}\left(\ln \left|w^2+2w+2\right|-2\arctan \left(w+1\right)\right)\right)\\\mathrm{Substitute\:back}\:w=\sqrt{v},\:v=2u,\:u=\tan \left(x\right)\\=\sqrt{2}\cdot \:2\left(\frac{1}{8}\left(\ln \left|\left(\sqrt{2\tan \left(x\right)}\right)^2-2\sqrt{2\tan \left(x\right)}+2\right|+2\arctan \left(\sqrt{2\tan \left(x\right)}-1\right)\right)-\frac{1}{8}\left(\ln \left|\left(\sqrt{2\tan \left(x\right)}\right)^2+2\sqrt{2\tan \left(x\right)}+2\right|-2\arctan \left(\sqrt{2\tan \left(x\right)}+1\right)\right)\right)\\=2\sqrt{2}\left(\frac{1}{8}\left(\ln \left|2\tan \left(x\right)-2\sqrt{2}\sqrt{\tan \left(x\right)}+2\right|+2\arctan \left(\sqrt{2}\sqrt{\tan \left(x\right)}-1\right)\right)-\frac{1}{8}\left(\ln \left|2\tan \left(x\right)+2\sqrt{2}\sqrt{\tan \left(x\right)}+2\right|-2\arctan \left(\sqrt{2}\sqrt{\tan \left(x\right)}+1\right)\right)\right)\\=2\sqrt{2}\left(\frac{1}{8}\left(\ln \left|2\tan \left(x\right)-2\sqrt{2}\sqrt{\tan \left(x\right)}+2\right|+2\arctan \left(\sqrt{2}\sqrt{\tan \left(x\right)}-1\right)\right)-\frac{1}{8}\left(\ln \left|2\tan \left(x\right)+2\sqrt{2}\sqrt{\tan \left(x\right)}+2\right|-2\arctan \left(\sqrt{2}\sqrt{\tan \left(x\right)}+1\right)\right)\right)+C\\$