Question

The value of ${\int }_0^{\pi /2}\frac{dx}{5+4\cos x}$ is

• A. $\frac{2}{3}{\tan }^{-1}\left(\frac{1}{3}\right)$
• B. ${\tan }^{-1}\left(\frac{1}{3}\right)$
• C. $\frac{5\pi }{4}$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{2}{3}{\tan }^{-1}\left(\frac{1}{3}\right)$

Let $I={\int }_0^{\pi /2}\frac{dx}{5+4\cos x}$

Now, $\cos x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}$

so, $\frac{1}{5+4\cos x}=\frac{1-{\tan }^2\left(x/2\right)}{9+{\tan }^2\left(x/2\right)}=\frac{{\sec }^2\left(x/2\right)}{9+{\tan }^2\left(x/2\right)}$

$I={\int }_0^{\pi /2}\frac{{\sec }^2\left(x/2\right)}{9+{\tan }^2\left(x/2\right)}dx$

Let $z=\tan \left(x/2\right)⟹2dz={\sec }^{2}xdx$

So, $I={\int }_0^{\pi /2}\frac{2}{9+z^2}dx$

$I=\left[\frac{2}{3}{\tan }^{-1}\right(z/3){]}_0^1$

So, $I=\frac{2}{3}{\tan }^{-1}\left(1/3\right)$