1 problem found
By making the change of variable \(t=\pi-x\) in the integral \[ \int_{0}^{\pi}x\mathrm{f}(\sin x)\,\mathrm{d}x, \] or otherwise, show that, for any function \(\mathrm{f},\) \[ \int_{0}^{\pi}x\mathrm{f}(\sin x)\,\mathrm{d}x=\frac{\pi}{2}\int_{0}^{\pi}\mathrm{f}(\sin x)\,\mathrm{d}x\,. \] Evaluate \[ \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x\quad\mbox{ and }\quad\int_{0}^{2\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x\,. \]
Solution: \begin{align*} && I &= \int_0^{\pi} x f(\sin x) \d x \\ t = \pi - x, \d t = -\d t : &&&= \int_{t = \pi}^{t = 0} (\pi - t) f(\sin (\pi - t)) -\d t \\ &&&= \int_0^{\pi} (\pi - t) f(\sin t) \d t \\ \Rightarrow && 2 I &= \pi \int_0^\pi f(\sin t) \d t \\ \Rightarrow && I &= \frac{\pi}{2} \int_0^{\pi} f(\sin x) \d x \end{align*} \begin{align*} && I &= \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} \d x \\ &&&= \frac{\pi}{2}\left [ -\tan^{-1} \cos x\right]_0^{\pi} \\ &&&= \tan 1 - \tan (-1) = \frac{\pi^2}{4} \\ \\ && I &= \int_{0}^{2\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ &&&= \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x + \int_{\pi}^{2\pi} \frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ u = x - \pi, \d u = \d x: &&&= \frac{\pi^2}{4} + \int_{0}^{\pi} \frac{(u+\pi)(-\sin u)}{1 + \cos^2 u}\d u \\ &&&= \frac{\pi^2}{4} -\frac{3\pi}{2} \int_0^{\pi} \frac{\sin u}{1+\cos^2 u} \d u \\ &&&= - \frac{\pi^2}2 \end{align*}