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Here's what I tried:

$$ \int \frac{\cos^2x-\sin^2x}{\sin x(1+\sin^2 x)}dx = \int \frac{\cos^2x}{\sin x (1+\sin^2x)}dx - \int \frac{\sin^2x}{\sin x (1+\sin^2x)}dx $$ I tried then to divide with $ \sin x $ and $ \cos x $. Tried to use some trigonometric identities, but it didn't work, I just complicated it more.

And I can't see something that I can substitute.

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  • $\begingroup$ Did you consider that you can cancel one "sin(x)" in the second integral ? $\endgroup$ – Peter Oct 16 '16 at 15:28
  • $\begingroup$ The substitution $t=\tan(\frac{x}{2})$ will work, but the resulting rational function could be difficult to integrate. $\endgroup$ – Peter Oct 16 '16 at 15:30
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    $\begingroup$ Yes, I tried to cancel one $\sin x$. $\endgroup$ – Gjekaks Oct 16 '16 at 15:33
  • $\begingroup$ An usual idea, but have you tried $t=\sin^2(x)$ ? $\endgroup$ – Peter Oct 16 '16 at 15:34
  • $\begingroup$ This expression could be integrated very easily! But only without $\sin$ and $\cos$ , of course. $\endgroup$ – Peter Oct 16 '16 at 15:36
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Hint. One may write, with the change of variable $u=\cos x$, $$ \int \frac{\cos^2x}{\sin x (1+\sin^2x)}dx=\int \frac{\cos^2x\:\sin x}{\sin^2 x (1+\sin^2x)}dx=-\int \frac{u^2}{(1-u^2) (2-u^2)}du $$ and $$ \int \frac{\sin^2x}{\sin x (1+\sin^2x)}dx=\int \frac{\sin x}{ 1+\sin^2x}dx=-\int \frac{du}{2-u^2} $$ then the new integrals are easier to evaluate.

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