What is the antiderivative of #cot(x)#?

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Answer 1 · 2018-03-24T03:01:37

#intcotxdx=ln|sinx|+C#

Recall that #cotx=cosx/sinx.# Thus,

#intcotxdx=intcosx/sinxdx#

We can solve this with a simple substitution.

#u=sinx#

#du=cosxdx#

This appears in our numerator, so the substitution is indeed valid.

So, we have

#int(du)/u=ln|u|+C=ln|sinx|+C#

#intcotxdx=ln|sinx|+C#