Integrating $\int\sin^{-2}xdx$

Question

I am trying to prove that $$ \int\frac{1}{\sin^2(x)}dx = -\cot(x) + C $$ but I have difficulties, I don't know where to start, I can't substitute anything with $sin(x)$ because I don't have a $cos(x)$ to make it up for, I also know that: $$ \sin^2x = \frac{1 - \cos(2x)}{2} $$ so I basically have: $$ \int\frac{2}{1 - \cos(2x)}dx $$ and there I am stuck, how do I proceed?

Answer 1

Note that $\sin x=\tan x\cos x$, and let $\tan x=t.$

Then, you'll have $$\begin{align}\int\frac{dx}{\sin^2x}&=\int\frac{1}{\tan^2x}\cdot\frac{dx}{\cos^2x}\\&=\int\frac{1}{t^2}dt\\&=-\frac 1t+C\\&=-\frac{1}{\tan x}+C\\&=-\cot x+C.\end{align}$$

Answer 2

We know that $ \frac{1}{sin x} = csc x $ then the integral will be $\int{csc^2 x }$ remember that the derivative of $cot x = -csc^2 x$ this is it: there you have it:

$$\int{csc^2 x} = - cot x + c$$

Answer 3

You can use the tangent half angle substitution $t = \tan \frac{x}{2}$ which yields $\sin(x) = \frac{2 t}{1+t^2}$ and ${\rm d}x = \frac{2}{1+t^2}{\rm d} t$.

$$ \int \frac{1}{\sin^2 x}\,{\rm d}x = \int \frac{1}{\left(\frac{2 t}{1+t^2}\right)^2} \left( \frac{2}{1+t^2} \right)\,{\rm d}t =\int \frac{1+t^2}{2 t^2} {\rm d}t = -\frac{1-t^2}{2 t}+C$$

Since $\sin x = \frac{2 t}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$ then $\tan x = \frac{2 t}{1-t^2}$.

So $$\frac{1-t^2}{2 t} = \frac{\cos x}{\sin x} = \cot x$$

and

$$ \int \frac{1}{\sin^2 x}\,{\rm d}x = -\cot x + C $$