Question

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

Answer 1

`x/8-1/32sin(4x)+C`

Explanation:

This can be written as:

`I=intcos^2(x)sin^2(x)dx`

We can rewrite this using the identity `sin(2x)=2sin(x)cos(x)`. Thus, `sin(x)cos(x)=sin(2x)/2`. Square both sides to see that `cos^2(x)sin^2(x)=sin^2(2x)/4`. Thus:

`I=1/4intsin^2(2x)dx`

Let `u=2x` so that `du=2dx`.

`I=1/8intsin^2(2x)*(2dx)=1/8intsin^2(u)du`

The way to integrate this is to use another double-angle formula.

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

Thus:

`I=1/16int(1-cos(2u))du`

`I=1/16intdu-1/16intcos(2u)du`

The first integral is the most basic integral and the second can be solved through inspection or substitution, where `v=2u`.

`I=1/16u-1/32sin(2u)+C`

Since `u=2x`:

`I=x/8-1/32sin(4x)+C`