How to use substitution to evaluate $ \int \frac{3}{\sin^2{x}} dx~ $?

Question

I've learned a couple of methods of integrating, but I'm still not sure when to use which one.

$$ \int \frac{3}{\sin^2{x}} dx $$

I tried using a method where I set something to $u$ and $dv$ and go from there, but I don't end up anywhere with this problem.
I know you can use substitution method and then integrate by parts, but I'm not sure which part of the integral I should begin substituting.

Answer 1

Just let $u = \frac{\cos x}{\sin x}$ and find $\frac{du}{dx}$ then proceed further.

Answer 2

Generally speaking, this is a standard integral

$$I = \int \frac{3}{\sin^2x}dx = 3\int\frac{1}{\sin^2x}dx = 3\int\frac{\sin^2x + \cos^2x}{(\sin x)^2}dx$$

Now if $v = \frac{\cos x}{\sin x} \implies \frac{dv}{dx} = ?$

Answer 3

Hint: recall that $(\cot x)'=-\frac{1}{sin^2 x}$

Answer 4

Question $\bf1$: I've learned a couple of methods of integrating, but I'm still not sure when to use which one.
Answer: Yes, there are so many methods to solve the integrations. And if you want to use the most effective one, you have to watch the integrand function carefully and think about the rules you learn. Basically the more you practice, more you learn.

Solution of the given integral : $$I=\int \frac{3}{\sin^2{x}} dx=3\int \dfrac{1/\cos^2(x)}{\sin^2(x)/\cos^2(x)} dx=3\int \dfrac{\sec^2(x)}{\tan^2(x)} dx$$Putting $~u=\tan(x)~\implies du=\sec^2(x)~ dx$,

$$I=3\int \dfrac{\sec^2(x)}{\tan^2(x)} dx=3\int \dfrac{du}{u^2}=-\dfrac 3u+c=-\dfrac{3}{\tan(x)}+c=-3\cot(x)+c$$where $~c~$ is arbitrary independent constant.

Answer 5

Another solution (even if the previous answers are simpler).

Use the tangent half-angle substitution $$I=\int \frac{3}{\sin^2{x}} dx=\frac{3}{2}\int \left(\frac{1}{t^2}+1\right)\,dt=\frac{3 t}{2}-\frac{3}{2 t}+C$$ Back to $x$ $$I=-3\cot(x)+C$$