Mathematics - Definite Integration Question with Solution | TestHub

We have, ${\int }_0^{\pi }\frac{1}{5+4\cos x}dx$ $={\int }_0^{\pi }\frac{1}{5+4\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}dx$ $={\int }_0^{\pi }\frac{1+{\tan }^2\frac{x}{2}}{5\left(1+{\tan }^2\frac{x}{2}\right)+4\left(1-{\tan }^2\frac{x}{2}\right)}dx$ $={\int }_0^{\pi }\frac{1+{\tan }^2\frac{x}{2}}{9+{\tan }^2\frac{x}{2}}\mathrm{dx}={\int }_0^{\pi }\frac{{\sec }^2\frac{x}{2}}{9+{\tan }^2\frac{x}{2}}\mathrm{dx}$ Let $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$ Also, $x=0\Rightarrow t=0$ and $x=\pi \Rightarrow t=\infty$ $\therefore \mathrm{I}={\int }_0^{\infty }\frac{\mathrm{dt}}{9+t^2}$ $\therefore \mathrm{I}=2{\int }_0^{\infty }\frac{\mathrm{dt}}{9+t^2}$ $\therefore \mathrm{I}=\frac{2}{3}{\left[{\tan }^{-1}\frac{t}{3}\right]}_0^{\infty }=\frac{2}{3}\left[{\tan }^{-1}\infty -{\tan }^{-1}0\right]$ $=\frac{2}{3}\left(\frac{\pi }{2}-0\right)=\frac{\pi }{3}$