Find the integral $\int \frac{1}{x^2 \cdot \tan(x)} \ dx$

Question

This problem seems pretty tricky. I need to find the integral of

$$\int \dfrac{1}{x^2 \cdot \tan(x)} \ dx$$

Any help would be greatly appreciated!

Answer 1

$\int x^{-2}(\tan x)^{-1}=\int x^{-2}\cot x$

then by parts u=$\cot x$, dv=$x^{-2}$

$\int x^{-2}\cot x= -x^{-1}\cot x-\int (-x^{-1})(-\csc^2x)=\int x^{-1}(-\csc^2x)-x^{-1}\cot x )$

Thus $$ \int \frac {1}{x^2\tan x}= \frac {-\cot x}{x}-\int \frac {csc^2x}{x}+c$$

Yep, no closed form

Answer 2

I would use the decomposition of $\frac{1}{\tan x}$ into simple fractions:

$$\frac{1}{\tan x}=\frac{1}{x}+\sum_{n=1}^{\infty}\frac{2x}{x^2-n^2\pi^2}$$ So

$$\frac{1}{x^2\tan x}=\frac{1}{x^3}+2\sum_{n=1}^{\infty}\frac{1}{x(x^2-n^2\pi^2)}$$ and

$$\int\frac{dx}{x^2\tan x}=-\frac{1}{2x^2}+\sum_{n=1}^{\infty}\frac{\ln\frac{\left |x^2-n^2\pi^2 \right |}{x^2}}{n^2\pi^2}+\text{const}$$

This is of course a matter of taste, whether the form is closed or not.