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

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.

marked as duplicate by Rory Daulton, Daniel W. Farlow, Rolf Hoyer, Steven Stadnicki, egregApr 20 '15 at 23:46

• 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
• math.stackexchange.com/questions/296548/… – 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

\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