Problems

Solution:

1. \(\,\) \begin{align*} && \sec^2\left(\tfrac14\pi-\tfrac12 x\right) &= \frac{1}{\cos^2 \left(\tfrac14\pi-\tfrac12 x\right)} \\ &&&= \frac{1}{\frac{1+\cos 2\left(\tfrac14\pi-\tfrac12 x\right)}{2}} \\ &&&= \frac{2}{1 + \cos \left(\tfrac12\pi- x\right)} \\ &&&= \frac{2}{1+\sin x} \\ \\ && \int \frac{1}{1+\sin x} \d x &= \int \tfrac12\sec^2\left(\tfrac14\pi-\tfrac12 x\right) \d x\\ &&&= - \tan\left(\tfrac14\pi-\tfrac12 x\right) + C \end{align*}
2. \(\,\) \begin{align*} && I &= \int_0^{\pi} x f(\sin x) \d x \\ y = \pi - x, \d y = - \d x: &&&= \int_{y=\pi}^{y = 0} (\pi - y) f(\sin(\pi - y))(-1) \d y \\ &&&= \int_0^\pi (\pi - y) f(\sin y) \d y \\ &&&= \pi \int_0^\pi f(\sin y) \d y - I \\ \Rightarrow && I &= \frac{\pi}{2} \int_0^\pi f(\sin x) \d x \\ \\ \Rightarrow && \int_0^{\pi} \frac{x}{1 + \sin x} \d x &= \frac{\pi}{2} \int_0^{\pi} \frac{1}{1 + \sin x} \d x\\ &&&=\frac{\pi}{2} \left [- \tan\left(\tfrac14\pi-\tfrac12 x\right) \right]_0^{\pi} \\ &&&= \frac{\pi}{2} \left (-\tan (-\tfrac{\pi}{4}) + \tan \tfrac{\pi}{4} \right) \\ &&&= \pi \end{align*}
3. \(\,\) \begin{align*} && J &= \int_0^{\pi} \frac{2x^3-3\pi x^2}{(1+\sin x)^2} \d x \\ y = \pi - x: &&&= \int_0^{\pi} \frac{2(\pi-y)^3-3\pi (\pi - y)^2}{(1+\sin x)^2 } \d y \\ &&&= \int_0^{\pi} \frac{-2 y^3 + 3 \pi y^2 - \pi^3}{(1+ \sin x)^2}\\ &&&= -\pi^3 \int_0^{\pi} \frac{1}{(1 + \sin x)^2} \d x -J \\ \Rightarrow && J &= -\frac{\pi^3}{2} \int_0^{\pi} \frac{1}{(1 + \sin x)^2} \d x\\ &&&= -\frac{\pi^3}{2} \int_0^\pi \tfrac14 \sec^4\left(\tfrac14\pi-\tfrac12 x\right) \d x \\ &&&= -\frac{\pi^3}{8} \int_0^\pi \sec^2\left(\tfrac14\pi-\tfrac12 x\right)\left (1 + \tan^2\left(\tfrac14\pi-\tfrac12 x\right) \right) \d x \\ &&&= -\frac{\pi^3}{8} \left [-\frac23 \tan^3\left(\tfrac14\pi-\tfrac12 x\right) - 2 \tan\left(\tfrac14\pi-\tfrac12 x\right) \right]_0^{\pi} \\ &&&= -\frac{\pi^3}{8} \left (\frac43+4 \right) \\ &&&= -\frac{2\pi^3}{3} \end{align*}