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Find the integral

$$ \int_0^{\pi} \frac{x \sin x}{1+(\cos x)^2}$$

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  • 1
    $\begingroup$ Use the variable change $x=\pi-y$. $\endgroup$ – user 1591719 Feb 6 '13 at 20:20
  • $\begingroup$ @Chris'sister Would this help ? $\endgroup$ – Amr Feb 6 '13 at 20:20
  • $\begingroup$ @Amr: yes, since you get the the same integral in the right side with the sign changed. $\endgroup$ – user 1591719 Feb 6 '13 at 20:26
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Integrate by parts:

$$\begin{align}\int_0^{\pi} dx\: \frac{x \sin x}{1+(\cos x)^2} &= -\int_0^{\pi} d(\cos{x})\: \frac{x}{1+(\cos x)^2}\\ &= -[x \arctan{\cos{x}}]_0^{\pi} + \underbrace{\int_0^{\pi} dx \:\arctan{\cos{x}}}_{\mathrm{this} = 0} \\ &= \frac{\pi^2}{4} \end{align}$$

Keep in mind that I used the principal branch of the arctangent.

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  • $\begingroup$ That's a fast way (+1) $\endgroup$ – user 1591719 Feb 6 '13 at 20:45
  • $\begingroup$ Thanks, and you got (+1) from me for doing the integral another way. I think both of us exploited the symmetry, but came across it differently. $\endgroup$ – Ron Gordon Feb 6 '13 at 20:47
  • $\begingroup$ Right. It's a kind of easy problem often met in high school. $\endgroup$ – user 1591719 Feb 6 '13 at 20:52
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Solution 1
Let the variable change $x=\pi-y$, and then

$$I=\int_0^{\pi} \frac{x \sin x}{1+(\cos x)^2}\mathrm{dx}=-\int_{0}^{\pi} \frac{ y\sin y}{1+(\cos y)^2}\mathrm{dy}+\pi\int_0^{\pi}\frac{\sin y}{1+(\cos y)^2}\mathrm{dy}$$ $$I=-\frac{\pi}{2}\int_0^{\pi}\frac{(\cos x)'}{1+(\cos x)^2}\mathrm{dx}$$ $$I=\frac{\pi}{2}[-\arctan(\cos x)]_0^{\pi}=\frac{\pi^2}{4}$$

Solution 2 (the fast way)

We recall and employ the formula

$$\int_0^\pi xf(\sin x )\mathrm{dx}=\frac \pi2\int_0^\pi f(\sin x )\mathrm{dx}$$

that I used in another answer you may see here. Then $$\int_0^{\pi} \frac{x \sin x}{1+(\cos x)^2}\mathrm{dx}=-\frac{\pi}{2}\int_0^{\pi} \frac{(\cos x)'}{1+(\cos x)^2}\mathrm{dx}=\frac{\pi}{2}[-\arctan(\cos x)]_0^{\pi/2}=\frac{\pi^2}{4}$$ $\quad$

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  • $\begingroup$ Thanks,but i was specifically instructed to use other methods other than that... $\endgroup$ – Sabrina Ahmed Feb 6 '13 at 20:30
  • $\begingroup$ @SabrinaAhmed: well, you didn't specify that in your question. $\endgroup$ – user 1591719 Feb 6 '13 at 20:32
  • $\begingroup$ Sorry about that. But do you think there's another way? $\endgroup$ – Sabrina Ahmed Feb 6 '13 at 20:39
  • $\begingroup$ @SabrinaAhmed: then check rlgordonma's answer that approached the problem differently. $\endgroup$ – user 1591719 Feb 6 '13 at 20:44

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