Integral by Substitution: Indefinite Integral
How to solve the given indefinite integral by using the integral by substitution: 2 examples and their solutions.
Example∫ (2x - 1)8 dx
Instead of expanding (2x - 1)8 by hand,
let's solve this integral by substitution.
Set 2x - 1 = t.
Differentiate both sides.
The derivative of 2x - 1 is 2dx.
And the derivative of t is dt.
Derivative of an Implicit Function
To make dx in the given integral,
change 2dx = dt
to dx = [1/2]dt.
2x - 1 = t
dx = [1/2]dt
Put these into ∫ (2x - 1)8 dx.
Then (given) = ∫ t8 [1/2]dt.
Take 1/2 out from the integral.
Solve the integral.
Write the coefficient 1/2.
The integral of x8 is [1/9]x9.
Integral of a Polynomial
This is an indefinite integral.
So write +C.
[1/2]⋅[1/9] = 1/18
2x - 1 = t
So change t back to (2x - 1).
So
[1/18](2x - 1)9 + C
is the answer.
Example∫ sin5 x cos x dx
Set sin x = t.
Differentiate both sides.
The derivative of sin x is cos x dx.
Derivative of sin x
And the derivative of t is dt.
cos x dx is already in the given integral.
So you don't have to change cos x dx = dt.
sin x = t
cos x dx = dt
Put these into the given integral:
∫ sin5 x cos x dx.
Then (given) = ∫ t5 dt.
The integral of x5 is [1/6]x6.
This is an indefinite integral.
So write +C.
sin x = t
So change t back to sin x.
So
[1/6]sin6 x + C
is the answer.