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Bonus integration problem we got at class: Integrate $\frac {x \sin x}{1+\cos^2x}$ between $0$ and $\pi$

So the lecturer gave this problem. I tried this really hard but couldn't succeed. It doesn't give me any bonus points at class, he just gave it as a nice challenge and it got me quite frustrated.

Any help will be great!

Thank you.

Mike121
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  • What were you trying? There seems to be a natural 'first' step, though it gets messy quickly... – Steven Stadnicki Apr 20 '15 at 22:19
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    http://math.stackexchange.com/questions/296548/i-need-assistance-in-integrating-fracx-sin-x1-cos-x2 – Ron Gordon Apr 20 '15 at 22:20
  • I tried integration by parts for $xsinx$ as the $u'$, etc'. Didn't really work out because I couldn't figure out the extra integral. Tried all variations of integration by parts probably. – Mike121 Apr 20 '15 at 22:21
  • @Mike121 Copying and pasting what RonGordon said. So what part of this solution are you having trouble with: $\begin{align}\int_0^{\pi} dx: \frac{x \sin x}{1+(\cos x)^2} &= -\int_0^{\pi} d(\cos{x}): \frac{x}{1+(\cos x)^2}\ &= -[x \arctan{\cos{x}}]0^{\pi} + \underbrace{\int_0^{\pi} dx :\arctan{\cos{x}}}{\mathrm{this} = 0} \ &= \frac{\pi^2}{4} \end{align}$ ? – randomgirl Apr 20 '15 at 22:35

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$$ \begin{align} &\int_0^\pi\frac{x\sin(x)\,\mathrm{d}x}{1+\cos^2(x)}\tag{1}\\ &=\int_0^\pi\frac{(\pi-x)\sin(x)\,\mathrm{d}x}{1+\cos^2(x)}\tag{2}\\ &=\frac\pi2\int_0^\pi\frac{\sin(x)\,\mathrm{d}x}{1+\cos^2(x)}\tag{3}\\ &=\frac\pi2\left[-\arctan(\cos(x))\vphantom{\int}\right]_0^\pi\tag{4}\\[3pt] &=\frac\pi2\left[\frac\pi4+\frac\pi4\right]\tag{5}\\[3pt] &=\frac{\pi^2}4\tag{6} \end{align} $$ Explanation:
$(2)$: substitute $x\mapsto\pi-x$
$(3)$: average $(1)$ and $(2)$
$(4)$: substitute $u=\cos(x)$; apply arctan integral; undo substitution
$(5)$: evaluate
$(6)$: simplify

robjohn
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  • would the downvoter care to comment? – robjohn Apr 20 '15 at 23:28
  • They were frightened by your logo. – marty cohen Apr 21 '15 at 00:06
  • @martycohen: they just undownvoted and then redownvoted. I guess they saw that I had expanded the answer a bit, but then got frightened again. – robjohn Apr 23 '15 at 00:29
  • Oh well, you (and I) can afford it. Your 153k is awesome. I know that up and down votes are anonymous, but I think it would be nice to see if a large number of down votes come from one person. The identity of that person would still be unknown. – marty cohen Apr 23 '15 at 00:52
  • @martycohen: yeah, I don't care so much about the points, I just wonder if there is something I am missing. Voting is capricious. – robjohn Apr 23 '15 at 00:55
  • Agree. I have been occasionally downvoted. When I think it is unwarranted, I say so. – marty cohen Apr 23 '15 at 04:22