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.)
 A: If you want to be that perverse.  I learned a way to integrate a power of sine, so why not apply it to the $-2$ power?  Keep one factor of sine, convert all others to cosine, substitute $u=\cos x$.  If we do that here, we get
$$
\int\frac{dx}{\sin^2 x} = \int \frac{\sin x dx}{\sin^3 x}
=\int\frac{\sin x dx}{(1-\cos^2 x)^{3/2}}
=\int\frac{-du}{(1-u^2)^{3/2}} .
$$
Then we can evaluate this integral (somehow, maybe even a trig substitution) to get
$$
\int\frac{-du}{(1-u^2)^{3/2}} = \frac{-u}{\sqrt{1-u^2}} + C = \frac{- \cos x}{\sin x} + C
$$
A: Alright, we could attempt a Weierstrass substitution if that's the sort of thing you want. Let $t=\tan(\frac{x}{2})$. Thus $\csc(x)=\frac{1+t^{2}}{2t}$ and $dx=\frac{2dt}{1+t^{2}}$. Therefore we have the following:
$$\int \csc^{2}(x)dx=\int \frac{1+t^{2}}{2t}\cdot\frac{1+t^{2}}{2t}\cdot\frac{2dt}{1+t^{2}}=\int\frac{1+t^{2}}{2t^{2}}dt=\frac{1}{2}\int (t^{-2}+1) dt$$
$$=\frac{1}{2} \left[ \frac{-1}{t}+t\right]+C=\frac{t^{2}-1}{2t}+C=\frac{\tan^{2}(x/2)-1}{2\tan(x/2)}+C$$
Which leads to the given result by application of the double angle formula.   
Letting $u=\tan(x)$ works too. We get $\csc^{2}(x)=1+\frac{1}{u^{2}}=\frac{1+u^{2}}{u^{2}}$, and $\frac{du}{1+u^{2}}=dx$. Therefore the integral is
$$\int \csc^{2}(x)=\int \frac{1+u^{-2}}{1+u^{2}}du=\int \frac{1}{u^{2}}du=\frac{-1}{\tan(x)}+C=-\cot(x)+C$$
A: Dividing both numerator and denominator by $\cos^2(x)$:
$\displaystyle\int{\csc^2(x)dx} = \int \dfrac{1}{\sin^2}dx = \int \dfrac{\sec^2(x)}{\tan^2(x)}dx$
Using substitution $u = \tan(x)$, $du = \sec^2(x)dx$:
$\displaystyle\int{\csc^2(x)dx} = \int \dfrac{1}{u^2}du = -\dfrac{1}{u} +C = -\dfrac{1}{\tan(x)} + C =-\cot(x) + C$
A: Use the trig identity for $\text{csc}^2x$ and write the expression in terms of $\text{sin }x$ and $\text{cos }x$:
$$
\int{\text{csc}^2x\text{ }dx}=\int{(\text{cot}^2x+1)\text{ }dx}=\int{\text{cot}^2x\text{ }dx}+\int{dx}=\int{\frac{\text{cos}^2x}{\text{sin}^2x}\text{ }dx}+\int{dx}
$$
Prepare for integration by parts:
$$
u=\text{cos }x~~~~~~~~~~du=-\text{ sin }x\text{ }dx~~~~~~~~~~dv=\frac{\text{cos }x}{\text{sin}^2x}\text{ }dx~~~~~~~~~~v=-\frac{1}{\text{sin }x}
$$
Integrate by parts and simplify:
$$
\int{\frac{\text{cos}^2x}{\text{sin}^2x}\text{ }dx}+\int{dx}=-\frac{\text{cos }x}{\text{sin }x}-\int{\frac{-\text{ sin }x}{-\text{ sin }x}\text{ }dx}+\int{dx}=-\text{ cot }x+C
$$
A: 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,
