Integral of Cot^2(x)+Cot^4(x)

Question:

$\int (\cot^{2}x+\cot^{4}x)\mathrm{d}x$

Solution: 

In this problem we would be using an trigonometric identity: $\cot^{2}x+1 = \csc^{2}x$

$\int (\cot^{2}x+\cot^{4}x)\mathrm{d}x$

$=\int \cot^{2}x\cdot(\cot^{2}x+1)\mathrm{d}x$

$=\int (\cot^{2}x\cdot\csc^{2}x)\mathrm{d}x$

Let's assume $u=\cot x$

So, $\mathrm{d}u=-\csc^2x\mathrm{d}x$. Now:

$=\int (\cot^{2}x\cdot\csc^{2}x)\mathrm{d}x$

Replacing u , du :

$=-\int u^2\mathrm{d}u$

$=-\frac{u^3}{3}+c$

As, $u=\cot x$

$=-\frac{\cot^2x}{3}+c$