How do you integrate #int sin(x)^3 * cos(x) dx#?

1 Answer
May 1, 2016

Use a #u#-substitution to get #intsin^3xcosxdx=sin^4x/4+C#.

Explanation:

What we have in this integral is a function, #sinx#, and its derivative, #cosx#. That means the integral is solvable using a #u#-substitution:
Let #u=sinx->(du)/dx=cosx->du=cosxdx#
With this substitution, #intsin^3xcosxdx# becomes:
#intu^3du#

This new integral is easily evaluated using the reverse power rule:
#intu^3du=(u^(3+1))/(3+1)+C=u^4/4+C#

Because #u=sinx#, we can substitute to get a final answer of:
#intsin^3xcosxdx=sin^4x/4+C#