How do you evaluate the integral #int cottheta#?

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Answer 1 · 2017-01-23T08:57:52

#intcotxdx=ln|sinx|+C#

using the standard result :

#int(f'(x))/(f(x))dx=ln|f(x)|+C#

we have

#intcotxdx=intcosx/sinxdx#

now: #d/(dx)(sinx)=cosx#

so the result follows

#intcotxdx=ln|sinx|+C#