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How do you show

$$\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x} \,\mathrm{d}x = \frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \,\mathrm{d}x$$

without integrating by parts, but only using substitution?

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Why do you want to know, and why do you impose that restriction? – GEdgar Mar 24 '12 at 20:22
up vote 18 down vote accepted

For any integral of the form

$$I = \int_0^{\pi} x f(\sin x) \mathrm{d}x $$

since $\sin(\pi-x) = \sin(x)$, using substitution $x = \pi-u$, or $u=\pi- x$, one may reduce this integral to the following using

when $x=0, u=\pi$, and when $x=\pi, u=0$, and also $ \mathrm{d}x = \mathrm{-d}u$

$$ \begin{align*} I = \int_0^{\pi} (\pi - u) f(\sin u) \mathrm{d}u &= \pi \int_0^{\pi} f(\sin u) \mathrm{d}u - \int_{\pi}^{0} u f(\sin u) \mathrm{(-d}u) \\ &= \pi \int_0^{\pi} f(\sin u) \mathrm{d}u - \int_0^{\pi} u f(\sin u) \mathrm{d}u \\ &= \pi \int_0^{\pi} f(\sin u) \mathrm{d}u - I\\ 2I &= \pi \int_0^{\pi} f(\sin u) \mathrm{d}u\\ I & = \frac{\pi}{2}\int_0^{\pi} f(\sin u) \mathrm{d}u = \frac{\pi}{2}\int_0^{\pi} f(\sin x) \mathrm{d}x \tag{A} \end{align*} $$

Apply $(A)$ for $\displaystyle{f(\sin x) = \frac{\sin x}{2-\sin^2 x} = \frac{\sin x}{1+\cos^2 x}}$

Although it was not asked, to further notice what is the value of the integral

by substituting further $\cos x = t, -\sin x \mathrm{d}x = \mathrm{d}t \hspace{5pt} cos(\pi) = -1, cos(0) = 1$

$$ \begin{align*} \frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x} \,\mathrm{d}x &= \frac{\pi}{2} \int_{1}^{-1} \frac{\mathrm{-d}t}{1+t^2}\\ &= \frac{\pi}{2} \int_{-1}^{1} \frac{\mathrm{d}t}{1+t^2}\\ &= \frac{\pi}{2} \left( \tan^{-1}(1) - \tan^{-1}(-1) \right) \\ &= \frac{\pi}{2} \left( \frac{\pi}{4} - \frac{-\pi}{4} \right) = \frac{\pi}{2} \left(\frac{\pi}{2}\right) = \frac{{\pi}^2}{4}\\ \end{align*} $$

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+1. Nice and elegant. – user17762 Mar 24 '12 at 21:41
When making a substitution, don't forget to change the bounds of integration. – chharvey Mar 25 '12 at 7:24

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