How do you integrate #int x^2 cot2 x dx # using integration by parts?

1 Answer
Jim H
Mar 18, 2016

As a first step choose #u=x^2# and #dv = cotx dx#

Explanation:

This gets us #du=2xdx# and #v = 1/2ln(sinx)#,
so the integral is equal to

#1/2x^2ln(sinx) - int x ln(sinx) dx#

Writing the integral now will involve the polylogarithmic function. You can read more about that here.

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