Integrals Involving sin(x) with Odd Power

Integrals Involving sin(x) with Odd Power

Tutorial to find integrals involving odd powers of sin(x). Exercises with answers are at the bottom of the page.

In what follows, C is the constant of integration.

Example 1: Evaluate the integral

sin3(x) dx


Solution to Example 1:

The main idea is to rewrite the power of sin(x) as the product of a term with power 1 and a term with an even power. Example: sin3(x) = sin2(x) sin(x). Hence the given integral may be written as follows:

sin3(x) dx = sin2(x) sin(x) dx

=
(1 - cos2(x)) sin(x) dx

We now let u = cos(x), hence du/dx = -sin(x) or -du = sin(x)dx and substitute in the given intergral to obtain

sin3(x) dx = - (1 - u2) du

sin3(x) dx = (1/3) u3 - u + C

Substitute u by cos(x) to obtain

sin3(x) dx = (1/3)cos3(x) - cos(x) + C

Example 2: Evaluate the integral

sin5(x) dx


Solution to Example 2:

Rewrite sin5(x) as follows sin5(x) = sin4(x) sin(x). Hence the given integral may be written as follows:

sin5(x) dx = sin4(x) sin(x) dx

We now use the identity sin2(x) = 1 - cos2(x) to rewrite sin4(x) in terms of power of cos(x) and rewrite the given integral as follows:

sin5(x) dx = (1 - cos2(x))2 sin(x) dx

We now let u = cos(x), hence du/dx = -sin(x) or du = -sin(x)dx and substitute in the given intergral to obtain

sin5(x) dx = - (1 - u2)2 du

Expand and calculate the integral on the right

sin5(x) dx = - (u4 - 2u2 + 1) du

= -(1/5)u
5 + (2/3)u3 - u + C

and finally

sin5(x) dx = -(1/5)cos5(x) + (2/3)cos3(x) - cos(x) + C

Exercises: Evaluate the following integrals.

1. sin7(x)dx

2. sin9(x)dx



Answers to Above Exercises

1. (1/7)cos7(x) - (3/5)cos5(x) + cos3(x) - cos(x) + C

2. -(1/9)cos9(x) + (4/7)cos7(x) - (6/5)cos5(x) + (4/3)cos3(x) - cos(x) + C

More references on integrals and their applications in calculus.

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