Question

How do you find the antiderivative of `sin^2 (2x) cos^3 (2x) dx`?

Answer 1

We start with a u-subsitution, letting `u = 2x`. Then `du = 2dx` and `dx = (du)/2`.

`int sin^2ucos^3u (du)/2`

`1/2intsin^2ucos^3udu`

`1/2intsin^2ucos^2ucosudu`

`1/2intsin^2u(1- sin^2u)cosudu`

`1/2int(sin^2u - sin^4u)cosudu`

Now let `n = sinu`. Then `dn = cosudu` and `du = (dn)/cosu`.

`1/2int(n^2 - n^4)cosu * (dn)/cosu`

`1/2int n^2 -n^4 dn`

`1/2(1/3n^3 - 1/4n^4) + C`

`1/6n^3 - 1/8n^4 + C`

`1/6sin^3u - 1/8sin^4u + C`

`1/6sin^3(2x) - 1/8sin^4(2x) + C`

Hopefully this helps!