Question

What is the integral of `sin^5(x)`?

Answer 1

`intsin^5xdx=sin^2x-sinx-1/3sin^3x+C`

Explanation:

We want `intsin^5xdx`

Rewrite the integrand as `sinx(sin^2x)^2`:

`intsinx(sin^2x)^2dx`

Recall the identity `sin^2x+cos^2x=1`. The identity also tells us that

`sin^2x=1-cos^2x`

`intsinx(1-cos^2x)^2dx`

This can be solved using a simple substitution. Let

`u=cosx, du=-sinxdx, -du=sinxdx`

Rewrite the integral:

`-int(1-u^2)^2=-int(1-2u+u^2)`

Integrate:

`-int(1-2u+u^2)=-(u-u^2+1/3u^3)+C=u^2-u-1/3u^3+C`

Rewriting in terms of `x` yields

`intsin^5xdx=sin^2x-sinx-1/3sin^3x+C`