<s>[INST] Trigonometric Integration with Negative Exponents How do you integrate $\csc^4 x/\cot^2 x$? I know that this is the same as $\csc^4 x \cot^{-2} x$ and when you use techniques in trig integrals you end up with
$$\int \csc^2 x \csc^2 x \cot^{-2} x \,dx = \int \left(1 + \cot^2 x\right)\left(\cot^{-2} x\right)\csc^2 x \,dx.$$
Making the substitution $u = \cot x \Rightarrow du = -\csc^2 x \,dx$,
$$\begin{align}
\ldots{} &= -\int \left(1 + u^2 \right) u^{-2} \,du \\
&= -\int \left( u^{-2} + u^0 \right) \,du \\
&= -\int \left( u^{-2} + 1 \right) \,du \\
&= -\left[ -u^{-1} + u \right] \\
&= (1/\cot x) - \cot x \,.
\end{align}$$
Is this the final answer or did I just mess up?
 [/INST] Another approach is to use 
$\displaystyle\int\frac{\csc^{4}x}{\cot^{2}x}dx=\int\frac{1}{\sin^{4}x}\cdot\frac{\sin^{2}x}{\cos^{2}x}dx=\int\frac{1}{\sin^{2}x\cos^{2}x}dx=\int\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}dx$
$\displaystyle=\int\left(\frac{1}{\cos^{2}x}+\frac{1}{\sin^{2}x}\right)dx=\int(\sec^{2}x+\csc^{2}x)dx=\tan x-\cot x+C.$
</s>