Question

How do you find the integral of `(cosx)^2/(sinx)`?

Answer 1

Use `cos^2x/sinx = cscx-sinx`

Explanation:

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

`int sinx dx = -cosx +C`

`int cscx dx = -ln abs(cscx+cotx) +C` You can either memorize this untegral of the trick to getting it:

`int cscx dx = int cscx (cscx+cotx)/(cscx+cotx) dx`

`= - int (-csc^2x-cscx cotx)/(cscx+cotx)dx`

`= -int 1/u du = -ln abs u +C`

I think you can finish `int cos^2x/sinx dx` from here.