How do you integrate $int x^3 cot x dx$ using integration by parts?

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Answer 1 · 2016-11-28T18:49:42

You can't.

The antiderivative of $x^3cotx$ involves the Polylogarithm Function evaluated at imaginary values.

$I = intx^3cotx dx$

Let $u = x^3$ so $du = 3x^2 dx$ and

let $dv = cotx dx$ so $v = intcotx dx = ln abssinx$ .

$I = x^3lnabssinx - 3int x^2 ln abs sinx dx$

Let $u = x^2$ so $du = 2x dx$

let $dv = ln abs sinx dx$ so $v = 1/2i(x^2+Li_2(e^(2ix)))-xln(1-e^(2ix))+xlnabs sinx$

Where $Li_2(x) = sum_(k=1)^oo x^k/k^2$ $" "$ (known as the polylogarithm or Jonquiere's function.)

At this point I'll let you finish yourself. (Because I'm out of my depth.)

How do you integrate int x^3 cot x dx int x^3 cot x dx using integration by parts?