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Could someone help with the following integration: $$\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$$

So far I have done the following, but I am stuck:

I denoted $ y=-\cos x $ then: $$\begin{align*}&\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&= \arccos(-1) \arctan 1+\arccos 1 \arctan(-1) - \int^1_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\\&=\frac{\pi^2}{4}-\int^{1}_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\end{align*}$$

Then I am really stuck. Could someone help me?

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$$I=\int_0^{\pi} \frac{-x\sin x}{1+\cos^2 x}\,dx=\int_0^{\pi} \frac{(x-\pi)\sin x}{1+\cos^2 x}dx\quad(x\to \pi-x)$$

$$\Rightarrow I=\frac{\pi}{2}\int_0^{\pi}\frac{-\sin x}{1+\cos^2 x}\,dx$$

Let $t=\cos x:$

$$I=\frac{\pi}{2}\int_{-1}^{1}-\frac{1}{1+t^2}\,dt=-\frac{\pi^2}{4}$$

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  • $\begingroup$ Very nice. Much quicker than finding an indefinite integral of the original integrand first. $\endgroup$ – Pratyush Sarkar Mar 7 '13 at 1:36
  • $\begingroup$ Nice! Mathematica gives a mess: \begin{align*}\frac{1}{2}(-\pi^2+\pi(&2\sin^{-1}(\frac{(-1)^{1/8}}{2^{1/4}})+\,2 \sin^{-1}\frac{(-1)^{7/8}}{2^{1/4}})\\&- i (\ln((1+i)-i \sqrt{2})-\ln((1-i)+i\sqrt{2})\\ &+\ln((1+i)+i \sqrt{2})-\ln(-i ((1+i)+\sqrt{2}))))\\& -2 i (\sin^{-1}\frac{(-1)^{1/8}}{2^{1/4}}) (-\ln((1-i)+i \sqrt{2})+\ln(-i ((1+i)+\sqrt{2})))\\& +\sin^{-1}(\frac{(-1)^{7/8}}{2^{1/4}})\\&(\ln(1-\frac{1+i}{\sqrt{2}})-\ln(1+\frac{1+i}{\sqrt{2}})\\& +\ln((1-i)+i \sqrt{2})-\ln((1+i)+i \sqrt{2})\\& -\ln(-i ((1+i)+\sqrt{2}))\\&+\ln(-\frac{(1-i) ((1+i)+\sqrt{2})}{-2+\sqrt{2}}))))\end{align*} $\endgroup$ – Steve Kass Mar 7 '13 at 5:03
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Let $$I = \int_0^{\pi} \dfrac{x \sin(x)}{1+\cos^2(x)} dx = \int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \sin(x+\pi/2)}{1 + \cos^2(x+\pi/2)} dx = \int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \cos(x)}{1 + \sin^2(x)} dx $$ Now $$\int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \cos(x)}{1 + \sin^2(x)} dx = \int_{-\pi/2}^{\pi/2} \underbrace{\dfrac{x \cos(x)}{1 + \sin^2(x)}}_{\text{Odd function}} dx + \dfrac{\pi}2 \cdot \int_{-\pi/2}^{\pi/2} \dfrac{\cos(x)}{1 + \sin^2(x)} dx $$ Hence, we get that $$I = \dfrac{\pi}2 \cdot \int_{-\pi/2}^{\pi/2} \dfrac{\cos(x)}{1 + \sin^2(x)} dx = \dfrac{\pi}2 \cdot \int_{-1}^1 \dfrac{dt}{1+t^2} = \dfrac{\pi}2 \cdot \left( \dfrac{\pi}4 - \dfrac{-\pi}4\right) = \dfrac{\pi^2}4$$ The integral you are after is $-I$ and hence the answer is $-\dfrac{\pi^2}4$.

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