Question

Integral of cot^4/cot^2x

Answer 1

$\bf 1+cot^2(\theta)=csc^2(\theta)\implies cot^2(\theta)=csc^2(\theta)-1\\\\ -------------------------------\\\\ \displaystyle \int~\cfrac{cot^4(x)}{cot^2(x)}\cdot dx\implies \int~\cfrac{[cot(x)]^4}{[cot(x)]^2}\cdot dx\implies \int~[cot(x)]^2\cdot dx \\\\\\ \displaystyle \int~cot^2(x)dx\implies \int~[csc^2(x)-1]dx\implies \int~csc^2(x)dx-\int1\cdot dx \\\\\\ -cot(x)-x+C$

Answer 2

$\int \frac{\cot^4x}{\cot^2x}dx = \int \cot^2x dx$

using the identity: $1+\cot^2x = \csc^2x$
$\int \cot^2x dx = \int( \csc^2x - 1 )dx$
$=-\cot x-x +c$