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

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Answer 1 · 2016-08-11T17:31:48

#-ln(abs(cscx+cotx))+cosx+C#

We have:

#intcos^2x/sinxdx#

Through the Pythagorean identity:

#=int(1-sin^2x)/sinxdx#

Split the integral:

#=intcscxdx-intsinxdx#

These are both well-known integrals:

#=-ln(abs(cscx+cotx))+cosx+C#