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

Nov 28, 2016

You can't.

#### Explanation:

The antiderivative of ${x}^{3} \cot x$ involves the Polylogarithm Function evaluated at imaginary values.

$I = \int {x}^{3} \cot x \mathrm{dx}$

Let $u = {x}^{3}$ so $\mathrm{du} = 3 {x}^{2} \mathrm{dx}$ and

let $\mathrm{dv} = \cot x \mathrm{dx}$ so $v = \int \cot x \mathrm{dx} = \ln \left\mid \sin \right\mid x$.

$I = {x}^{3} \ln \left\mid \sin \right\mid x - 3 \int {x}^{2} \ln \left\mid \sin \right\mid x \mathrm{dx}$

Let $u = {x}^{2}$ so $\mathrm{du} = 2 x \mathrm{dx}$

let $\mathrm{dv} = \ln \left\mid \sin \right\mid x \mathrm{dx}$ so $v = \frac{1}{2} i \left({x}^{2} + L {i}_{2} \left({e}^{2 i x}\right)\right) - x \ln \left(1 - {e}^{2 i x}\right) + x \ln \left\mid \sin \right\mid x$

Where $L {i}_{2} \left(x\right) = {\sum}_{k = 1}^{\infty} {x}^{k} / {k}^{2}$ $\text{ }$ (known as the polylogarithm or Jonquiere's function.)

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