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

1 Answer
Oct 11, 2015

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.