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

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Answer 1 · 2015-10-11T22:30:40

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

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

$int sinx dx = -cosx +C$

$int cscx dx = -ln abs(cscx+cotx) +C$ 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.

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