Integral of cosec squared ($\operatorname{cosec}^2x$, $\csc^2x$) According to my sheet of standard integrals,
$\int \csc^2x \, dx = -\cot x + C$.
I am interested in a proof for the integral of $\operatorname{cosec}^2x$ that does not require differentiating $\cot x$. (I already know how to prove it using differentiation, but am interested in how one would calculate it without differentiation.)

Question: $\displaystyle\int{\csc^2(x)dx} = $
Using substitution: $u = \dfrac{1}{\sin^2(x)}$ 
$\dfrac{du}{dx} = \dfrac{-2\cos(x)}{\sin^3(x)} \quad \longrightarrow du = \left(\dfrac{-2\cos(x)}{\sin(x)}\right)\times \left(\dfrac{1}{\sin^2(x)}\right)dx$
$\therefore -\dfrac{1}{2}\tan(x)du = \dfrac{1}{\sin^2(x)} dx \qquad (1)$
We still must find what is $\tan(x)$ in terms of $u$, hence:
$u = \dfrac{\sin^2(x) + \cos^2(x)}{\sin^2(x)} = 1+ \cot^2(x)$
$u-1 = \cot^2(x)$
$\sqrt{u-1} = \cot(x)$
$\dfrac{1}{\sqrt{u-1}} = \tan(x)$
Therefore substitute in $(1)$ and our integral becomes:
$-\dfrac{1}{2}\displaystyle\int{(u-1)^{-\frac{1}{2}}}du = -\dfrac{1}{2}\dfrac{\sqrt{(u-1)}}{\dfrac{1}{2}} + C = -\sqrt{\left(\dfrac{1}{\sin^2(x)} - 1)\right)} + C $
$= -\sqrt{\left(\dfrac{1-\sin^2(x)}{\sin^2(x)}\right)} + C = -\cot(x) + C$
Hope this makes sense,
