Paragraph: Let I_1=∫_0^π (x)/(1+ ^2 x) d x and I_2=∫_0^π (1)/(1+ ^2...

\quad I_1=\int_0^\pi \frac{x}{1+\cos ^2 x} d x$....(1)$I_1=\int_0^\pi \frac{(\pi-x)}{1+\cos ^2(\pi-x)} d x$....(2) (1) + (2),$2 I_1=\int_0^\pi \frac{\pi}{1+\cos ^2 x} d xI \int \frac{\sec }{1 \sec } d xI_1=\frac{\pi}{2} \int_0^\pi \frac{\sec ^2 x}{2+\tan ^2 x} d x[\because f(\pi-x)=f(x)]I_1=\frac{\pi}{2} \cdot 2 \int_0^{\pi / 2} \frac{\sec ^2 x d x}{2+\tan ^2 x}$Let$\tan x=t, \sec ^2 x d x=d tI_1=\pi \int_0^{\infty} \frac{d t}{t^2+(\sqrt{2})^2}I_1=\left.\frac{\pi}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)\right|_0 ^{\infty}=\frac{\pi}{\sqrt{2}}\left(\frac{\pi}{2}-0\right)=\frac{\pi^2}{2 \sqrt{2}}$Similarly,$I_2=\int_0^\pi \frac{1}{1+\sin ^2 x} d x=2 \int_0^{\pi / 2} \frac{d x}{1+\sin ^2 x}=2 \int_0^{\pi / 2} \frac{d x}{2 \sin ^2 x+\cos ^2 x}=2 \int_0^{\pi / 2} \frac{\sec ^2 x d x}{2 \tan ^2 x+1}=\int_0^{\pi / 2} \frac{\sec ^2 x d x}{\tan ^2 x+\left(\frac{1}{\sqrt{2}}\right)^2} Let $\tan x=t, \sec ^2 x d x=d t\int_0^{\infty} \frac{d t}{t^2+\left(\frac{1}{\sqrt{2}}\right)^2}=\left.\sqrt{2} \tan ^{-1} \sqrt{2} t\right|_0 ^{\infty}=\sqrt{2}\left(\frac{\pi}{2}-0\right)=\frac{2 \pi}{2 \sqrt{2}}\frac{I_1+I_2}{I_1-I_2}=\frac{\frac{\pi^2}{2 \sqrt{2}}+\frac{2 \pi}{2 \sqrt{2}}}{\frac{\pi^2}{2 \sqrt{2}}-\frac{2 \pi}{2 \sqrt{2}}}=\frac{\pi^2+2 \pi}{\pi^2-2 \pi}\Rightarrow \frac{I_1+I_2}{I_1-I_2}=\frac{\pi+2}{\pi-2}$