# Question d405a

Nov 24, 2017

#### Explanation:

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Nov 24, 2017

${\int}_{0}^{\pi} \frac{x \sin x}{1 + {\left(\cos x\right)}^{2}} \cdot \mathrm{dx} = {\pi}^{2} / 4$

#### Explanation:

$I = {\int}_{0}^{\pi} \frac{x \sin x}{1 + {\left(\cos x\right)}^{2}} \cdot \mathrm{dx}$

After using $x = \pi - y$ an $\mathrm{dx} = - \mathrm{dy}$ transforms, $I$ became

$I = {\int}_{\pi}^{0} \frac{\left(\pi - y\right) \cdot \sin \left(\pi - y\right) \left(- \mathrm{dy}\right)}{1 + {\left(\cos \left(\pi - y\right)\right)}^{2}}$

=${\int}_{\pi}^{0} \frac{- \left(\pi - y\right) \cdot \sin y \cdot \mathrm{dy}}{1 + {\left(- \cos y\right)}^{2}}$

=${\int}_{\pi}^{0} \frac{- \left(\pi - y\right) \cdot \sin y \cdot \mathrm{dy}}{1 + {\left(\cos y\right)}^{2}}$

=${\int}_{0}^{\pi} \frac{\left(\pi - y\right) \cdot \sin y}{1 + {\left(\cos y\right)}^{2}} \cdot \mathrm{dy}$

=${\int}_{0}^{\pi} \frac{\left(\pi - x\right) \cdot \sin x}{1 + {\left(\cos x\right)}^{2}} \cdot \mathrm{dx}$

After collecting 2 integrals,

$2 I = {\int}_{0}^{\pi} \frac{x \sin x}{1 + {\left(\cos x\right)}^{2}} \cdot \mathrm{dx} + {\int}_{0}^{\pi} \frac{\left(\pi - x\right) \cdot \sin x}{1 + {\left(\cos x\right)}^{2}} \cdot \mathrm{dx}$

=$\pi \cdot {\int}_{0}^{\pi} \sin \frac{x}{1 + {\left(\cos x\right)}^{2}} \cdot \mathrm{dx}$

=$\pi \cdot {\left[- \arctan \left(\cos x\right)\right]}_{0}^{\pi}$

=$\left(- \pi\right) \cdot \left[\arctan \left(\cos \left(\pi\right)\right) - \arctan \left(\cos 0\right)\right]$

=$\left(- \pi\right) \cdot \left[\arctan \left(- 1\right) - \arctan \left(1\right)\right]$

=$\left(- \pi\right) \cdot \left[\left(- \frac{\pi}{4}\right) - \left(\frac{\pi}{4}\right)\right]$

=$\left(- \pi\right) \cdot \left(- \frac{\pi}{2}\right)$

=${\pi}^{2} / 2$

Thus, $I = {\int}_{0}^{\pi} \frac{x \sin x}{1 + {\left(\cos x\right)}^{2}} \cdot \mathrm{dx} = {\pi}^{2} / 4$