Question

The value of ${\int }_0^{\pi /2}\frac{d\theta }{5+3\cos \theta }$ is?

• A. ${\tan }^{-1}\frac{1}{2}$
• B. ${\tan }^{-1}\frac{1}{3}$
• C. $\frac{1}{2}{\tan }^{-1}\frac{1}{2}$
• D. $\frac{1}{3}{\tan }^{-1}\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{2}{\tan }^{-1}\frac{1}{2}$

Let $I={\int }_0^{\pi /2}\frac{d\theta }{5+3\cos \theta }={\int }_0^{\pi /2}\frac{d\theta }{5+3.\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}}$

$={\int }_0^{\pi /2}\frac{{\sec }^2\frac{\theta }{2d\theta }}{8+2{\tan }^2\frac{\theta }{2}}$

$={\int }_0^{\pi /2}\frac{\frac{1}{2}{\sec }^2\left(\frac{\theta }{2}\right)d\theta }{4+{\tan }^2\left(\frac{\theta }{2}\right)}={\int }_0^1\frac{dt}{4+t^2}$

$[$Put $\tan \frac{\theta }{2}=t$

So, $\frac{1}{2}{\sec }^2\frac{\theta }{2}d\theta =dt]$

$\therefore I==\frac{1}{2}{\left[{\tan }^{-1}\frac{t}{2}\right]}_0^1$

$=\frac{1}{2}({\tan }^{-1}\frac{1}{2}-{\tan }^{-1}0)$

$=\frac{1}{2}{\tan }^{-1}\frac{1}{2}$