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I try to evaluating $\int \:x\csc \left(x^2\right)\cot \left(x^2\right)dx$

let $u=x^2,\quad \quad du=2xdx,\:\quad \:dx=\frac{1}{2x}du$

then i get

$\int \:x\csc \left(u\right)\cot \left(u\right)\frac{1}{2x}du$ $=\frac{1}{2}\int \:\csc \left(u\right)\cot \left(u\right)du$

I'm stuck here, I've tried to evaluate the integrand but it seems like I can't find the right solution to continue finishing this problem. Please anyone help me, I would be really happy if you want to give me an advice to make this problem clear.

I know the answer would be (according to table of integral)

$-\frac{1}{2}\frac{1}{\sin \left(x^2\right)}+C$

but I want to know how to find the answer from scratch. >.<

EDIT :

thank you everyone, finally i found the right answer through

$v\:=\:csc\left(u\right)$ then $dv\:=\:-csc\left(u\right)cot\left(u\right)du$

so,

$=-\frac{1}{2}\int dv\:=\:-\frac{v}{2}\:+\:C\:$

substitute back u and v then i get

$-\frac{1}{2}\frac{1}{\sin \left(x^2\right)}+C$

Thank you so much for your help & advice. :)

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    $\begingroup$ You were never taught the derivative of cosecant? $\endgroup$ – Mike Sep 2 '14 at 2:36
  • $\begingroup$ Sorry, no one ever taught me, I just learned it by myself. This is not for homework or anything else that has something to do with school. It's just my curiosity to knowing the method behind Math problems. :) $\endgroup$ – alethiologist Sep 2 '14 at 2:41
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Hint: $(\csc(u))'=-(\csc(u)\cot(u))$

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Let $u=\csc(x^{2})$. Then $du=-2x\csc(x^{2})\cot(x^{2})dx$. Once you account for the $-2$ that has appeared in the $du$ portion, you will have it.

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In case you do not know the anti-derivative $\csc \left(u\right)\cot \left(u\right)$ advance as

$$ I = \frac{1}{2}\int \:\csc \left(u\right)\cot \left(u\right)du = \frac{1}{2}\int \:\frac{1}{\sin(x)}\frac{\cos(u)}{\sin(u)}du$$

$$=\frac{1}{2}\int \:(\sin(u))^{-2}{\cos(u)}du = -\frac{1}{2}(\sin(x))^{-1}$$

$$=-\frac{1}{2}\frac{1}{\sin(x)}= -\frac{1}{2}\csc(x). $$

You can use substitution to finish the problem.

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