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

The main content of this page would be the integration of functions using the indirect substitution method.

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 ∫sin^px dx, ∫cos^qx dx (p less than or equal to 4), sin^px and cos^qx are expressed in terms of sines and cosines of multiples of x using the identities of trigonometry.

Steps to evaluate integrals of the form sin^px and cos^qxdx,

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

Solution:

We know,

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

= (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