Question

How to integrate `int sin^2t cos^4t dt`?

Answer 1

We could write it as `int (cos^4x-cos^6x)dx` then use power reduction formulas to integrate `cos^4x` and `cos^6x` separately.

It may be slightly simpler to rewrite as:

`int sin^2xcos^2xcos^2x dx = int (sinxcosx)^2cos^2x dx`

`= int(1/2(sin2x))^2 cos^2x dx`

Now expand the square in the first factor and use power reduction on `cos^2x` to get

`1/8intsin^2 2x(1+cos 2x) dx`

`= 1/8intsin^2 2x dx +1/8 int sin^2 2xcos 2x dx`

`intsin^2 2x dx` may be evaluated by power reduction and

`int sin^2 2xcos 2x dx` is a `u = sin 2x` substitution