Question

$$\underset{0}{\overset{\pi }{\int }}\frac{1}{5+3\cos x}dx$$

Answer 1

$$LetI={\int }_0^{\pi }\frac{1}{5+3\cos x}dx.Then,$$
$$I={\int }_0^{\pi }\frac{1}{5+3\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}dx$$
$$\Rightarrow I={\int }_0^{\pi }\frac{1+{\tan }^2\frac{x}{2}}{5+5{\tan }^2\frac{x}{2}-3{\tan }^2\frac{x}{2}}dx$$
$$\Rightarrow I={\int }_0^{\pi }\frac{{\sec }^2\frac{x}{2}}{5+2{\tan }^2\frac{x}{2}}dx$$
$$Let\tan \frac{x}{2}=t.Then,\frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$$
$$Whenx=0,t=0andx=\pi ,t=\mathrm{\infty }$$
$$\therefore I={\int }_0^{\mathrm{\infty }}\frac{dt}{5+2t^2}$$
$$\Rightarrow I=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{dt}{\frac{5}{2}+t^2}$$
$$\Rightarrow I=\frac{1}{2}{[{\tan }^{-1}\frac{\sqrt{5}t}{\sqrt{2}}]}_0^{\mathrm{\infty }}$$
$$\Rightarrow I=\frac{1}{2}(\frac{\pi }{2}-0)$$
$$\Rightarrow I=\frac{\pi }{4}$$