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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.

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  • $\begingroup$ Is it $\int \frac{1}{\text (sin)^{2}x } dx$? $\endgroup$ – 00GB Jan 21 at 3:14
  • $\begingroup$ Please use MathJax - as it stands the objective integral is not clear $\endgroup$ – Dhanvi Sreenivasan Jan 21 at 3:17
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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 $v = \frac{\cos x}{\sin x} \implies \frac{dv}{dx} = ?$

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  • $\begingroup$ Thanks,but how did you find cosx/sinx and set it to u ? $\endgroup$ – Luke Jan 21 at 3:28
  • $\begingroup$ Well, firstly I knew that $\frac{d}{dx} \cot x = \csc^2 x$, as part of the differentials of elementary trig functions $\endgroup$ – Dhanvi Sreenivasan Jan 21 at 3:59
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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.

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  • $\begingroup$ Thanks, that is a new way of thinking $\endgroup$ – Luke Jan 21 at 3:36
  • $\begingroup$ You are welcome. @Luke $\endgroup$ – nmasanta Jan 21 at 4:29
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Just let $u = \frac{\cos x}{\sin x}$ and find $\frac{du}{dx}$ then proceed further.

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  • $\begingroup$ Thanks,but how did you find cosx/sinx and set it to u ? $\endgroup$ – Luke Jan 21 at 3:28
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Hint: recall that $(\cot x)'=-\frac{1}{sin^2 x}$

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  • $\begingroup$ Thanks, it is informative! $\endgroup$ – Luke Jan 21 at 3:35
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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 $x$ $$I=-3\cot(x)+C$$

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