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How to calculate $$\int_0^\pi \frac{x\sin x}{1+\cos^2x}\ dx\ ?$$ I wish I could say I ran out of ideas, but actually I have none.

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Just make the change of variables $x\rightarrow\pi-x$ and add the two results: $$2I=\int_0^\pi \frac{x\sin x}{1+\cos^2x}\ dx+\int_0^\pi \frac{(\pi-x)\sin x}{1+\cos^2x}\ dx=\pi \int_0^\pi \frac{\sin x}{1+\cos^2x}dx.$$ I believe you can take it from here.

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  • $\begingroup$ Yeah, I just looked at this post and thought about that. Easy. $\endgroup$ – Jules May 19 '14 at 19:06
  • $\begingroup$ @Jules Yes, this is a rather standard trick. It remains to make a further change of variables $y=\cos x$ and you are done. $\endgroup$ – Start wearing purple May 19 '14 at 19:08
  • $\begingroup$ May the downvoter explain his problem with this answer? $\endgroup$ – Start wearing purple May 19 '14 at 19:42
  • $\begingroup$ Perhaps the downvoter doesn't realise why the last integral is easier (the Weierstrass substitution). $\endgroup$ – Bennett Gardiner May 20 '14 at 3:29
  • $\begingroup$ Actually, as O.L. noticed, you don't even need Weierstrass substitution. $\endgroup$ – Jules May 20 '14 at 15:57

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