Question

How do you find the indefinite integral of ∫cos^2 xdx / sin^4 x ?

Answer 1

`intcos^2x/sin^4xdx=-1/3cot^3x+C`

Explanation:

So, we want `intcos^2x/sin^4xdx`.

Recall the identity `cos^2x=1-sin^2x`, and rewrite:

`int(1-sin^2x)/sin^4xdx=intdx/sin^4x-intdx/sin^2x`

`intdx/sin^2x-intcsc^2xdx`

`intcsc^2xdx=-cotx`, this is a simple integral. Solve the other integral:

`intdx/sin^4x=intcsc^4xdx`

`intcsc^2xcsc^2xdx`

Recalling that `1+cot^2x=csc^2x,` rewrite:

`int(1+cot^2x)csc^2xdx`

This can be solved with a simple substitution.

`u=cotx`

`du=-csc^2xdx`

`-du=csc^2xdx`

We then have

`-int(1+u^2)du=-u-1/3u^3`

Rewriting in terms of `x` yields

`intdx/sin^4x=-cotx-1/3cot^3x`

Thus, combining the integrals we found,

`intcos^2x/sin^4xdx=cancel(-cotx)-1/3cot^3x+cancel(cotx)+C`

`intcos^2x/sin^4xdx=-1/3cot^3x+C`