Question

How do I evaluate the indefinite integral `intsin^3(x)*cos^2(x)dx` ?

Answer 1

The answer is `-(cos^3x)/3+(cos^5x)/5+C`.

The trick with sinusoidal powers is to use identities so that you can have `sin x` or `cos x` with a power of 1 and use substitution.

In this case, it is easier to get `sin x` to a power of 1 using `sin^2x=1-cos^2x`.

`int sin^3x*cos^2x dx`
`=int sin x(1-cos^2x)cos^2x dx`
`=int sin x(cos^2x-cos^4x)dx`
`=int sin x cos^2xdx-int sin x cos^4x dx`

Now it is a matter of using substitution:

`u=cos x`
`du = -sin x dx`

`int sin x cos^2xdx-int sin x cos^4x dx`
`=int -u^2 du+ int u^4 du`
`=-(u^3)/3+(u^5)/5+C`
`=-(cos^3x)/3+(cos^5x)/5+C`