Question

How do you find the integral of `sin^2(2x) dx`?

Answer 1

The answer is `=x/2-(sin4x)/8+C`

Explanation:

We use

`cos4x=1-2sin^2(x)`

`sin^2 2x=1/2(1-cos4x)`

Therefore,

`int(sin^2 2x)dx=1/2int(1-cos4x)dx`

`=1/2(x-(sin4x)/4)+C`

`=x/2-(sin4x)/8+C`