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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.