Question

If $0<\alpha <\pi /2$ then the value of ${\int }_0^{\alpha }\frac{dx}{1-\cos x\cos \alpha }$ is

• A. $\pi /\alpha$
• B. $\pi /2\sin \alpha$
• C. $\pi /2\cos \alpha$
• D. $\pi /2\alpha$

Answer 1

Answer: (B)

$\pi /2\sin \alpha$

Let $I={\int }_0^{\alpha }\frac{dx}{1-\cos x\cos \alpha }$
$={\int }_0^{\alpha }\frac{dx}{({\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2})-\cos \alpha ({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})}$
$={\int }_0^{\alpha }\frac{dx}{(1-\cos \alpha ){\cos }^2\frac{x}{2}+2{\cos }^2\frac{\alpha }{2}{\sin }^2\frac{x}{2}}$
$=\frac{1}{2}{\int }_0^{\alpha }\frac{{\sec }^2\frac{\alpha }{2}{\sec }^2\frac{x}{2}}{{\tan }^2\frac{\alpha }{2}+{\tan }^2\frac{x}{2}}dx$
Substitute $\tan \frac{x}{2}=t$
$I={\int }_0^{\tan \frac{\alpha }{2}}\frac{{\sec }^2\frac{\alpha }{2}}{{\tan }^2\frac{\alpha }{2}+t^2}dt={\sec }^2\frac{\alpha }{2}\cot \frac{\alpha }{2}{\left[{\tan }^{-1}\frac{1}{\tan \frac{\alpha }{2}}\right]}_0^{\tan \frac{\alpha }{2}}$
$=\frac{\pi }{2\sin \alpha }$