Integrals Involving sin(x) and cos(x) with Odd Power
Tutorial to find integrals involving the product of powers of sin(x) and cos(x) with one of the two having an odd power. Examples and exercises with solutions are included.
Examples with Detailed solutions
In what follows, C is the constant of integration.
Example 1
Evaluate the integral
sin3(x) cos2(x) dx
Solution to Example 1: The main idea is to rewrite the integral writing the term with the odd power 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) cos2(x) dx = sin2(x) cos2(x) sin(x) dx
We now use the identity sin2(x) = 1 - cos2(x) and rewrite the given integral as follows:
We now let u = cos(x), hence du/dx = -sin(x) or -du = sin(x)dx and substitute in the given integral to obtain sin3(x) cos2(x) dx = -(1 - u2) u2 du
Expand and calculate the integral on the right sin3(x) cos2(x) dx = u4 - u2 du
= (1/5)u5 - (1/3)u3 + C
Substitute u by cos(x) to obtain sin3(x) cos2(x) dx = (1/5)cos5(x) - (1/3)cos3(x) + C
Example 2
Evaluate the integral
sin12(x) cos5(x) dx
Solution to Example 2: Rewrite cos5(x) as follows cos5(x) = cos4(x) cos(x). Hence the given integral may be written as follows: sin12(x) cos5(x) dx
= sin12(x) cos4(x) cos(x) dx
We now use the identity cos2(x) = 1 - sin2(x) to rewrite cos4(x) in terms of power of sin(x) and rewrite the given integral as follows: sin12(x) cos5(x) dx
= sin12(x) (1 - sin2(x))2 cos(x) dx
We now let u = sin(x), hence du/dx = cos(x) or du = cos(x)dx and substitute in the given integral to obtain sin12(x) cos5(x) dx
= u12 (1 - u2)2 du
Expand and calculate the integral on the right sin12(x) cos5(x) dx
= u12 (1 + u4 - 2u2) du
= (u16 - 2u14 + u12 ) du
= (1/17)u17 - (2/15)u15 + (1/13)u13 + C
Substitute u by sin(x) to obtain sin3(x) cos2(x) dx = (1/17)sin17(x) - (2/15)sin15(x) + (1/13)sin13(x) + C
Exercises
Evaluate the following integrals.
1. cos3(x) sin2(x) dx
2. sin3(x) cos14(x) dx