Question

The value of ${\int }_0^{\pi }\frac{1}{5+4\cos x}\mathrm{dx}$ is

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is C $\frac{\pi }{3}$

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