Question

How do you integrate `int sin^2(2x)dx`?

Answer 1

`1/8(4-sin(4x))+C`

Explanation:

Use the cosine double-angle identity to rewrite the function:

`cos(2alpha)=1-2sin^2(alpha)" "=>" "sin^2(alpha)=(1-cos(2alpha))/2`

Then:

`sin^2(2x)=(1-cos(4x))/2`

So:

`intsin^2(2x)dx=1/2int(1-cos(4x))dx=1/2intdx-1/2intcos(4x)dx`

The second can be solved with the substitution `u=4x=>du=4dx`:

`intsin^2(2x)dx=1/2x-1/8int4cos(4x)dx=1/2x-1/8intcos(u)du`

The integral of cosine is sine:

`intsin^2(2x)dx=1/2-1/8sin(u)=1/8(4-sin(u))=1/8(4-sin(4x))+C`