Question

If $\underset{0}{\overset{\alpha }{\int }}\frac{dx}{1-\cos \alpha \cos x}=\frac{A}{\sin \alpha }+B(\alpha \ne 0),$ then possible values of $A$ and $B$ are

• A. $A=\frac{\pi }{2},B=0$
• B. $A=\frac{\pi }{4},B=\frac{\pi }{4\sin \alpha }$
• C. $A=\frac{\pi }{6},B=\frac{\pi }{\sin \alpha }$
• D. $A=\pi ,B=\frac{\pi }{\sin \alpha }$

Answer 1

Answer: (A), (B)

$A=\frac{\pi }{4},B=\frac{\pi }{4\sin \alpha }$

Let $I=\underset{0}{\overset{\alpha }{\int }}\frac{dx}{1-\cos \alpha \cos x}$
$=\underset{0}{\overset{\alpha }{\int }}\frac{dx}{{\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2}-\cos \alpha ({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})}$
$=\underset{0}{\overset{\alpha }{\int }}\frac{dx}{(1-\cos \alpha ){\cos }^2\frac{x}{2}+(1+\cos \alpha ){\sin }^2\frac{x}{2}}$
$=\underset{0}{\overset{\alpha }{\int }}\frac{dx}{2{\sin }^2\frac{\alpha }{2}{\cos }^2\frac{x}{2}+2{\cos }^2\frac{\alpha }{2}{\sin }^2\frac{x}{2}}$
$=\frac{1}{2}\underset{0}{\overset{\alpha }{\int }}\frac{{\sec }^2\frac{\alpha }{2}\cdot {\sec }^2\frac{x}{2}}{{\tan }^2\frac{\alpha }{2}+{\tan }^2\frac{x}{2}}dx$

Put $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
$\therefore I={\sec }^2\frac{\alpha }{2}\underset{0}{\overset{\tan \left(\alpha /2\right)}{\int }}\frac{dt}{t^2+{\tan }^2\frac{\alpha }{2}}$
$={\sec }^2\frac{\alpha }{2}\cdot \cot \frac{\alpha }{2}{\left[{\tan }^{-1}\left(\frac{t}{\tan \frac{\alpha }{2}}\right)\right]}_0^{\tan \left(\alpha /2\right)}$
$={\sec }^2\frac{\alpha }{2}\cdot \cot \frac{\alpha }{2}\cdot \frac{\pi }{4}$
$=\frac{\pi }{2\sin \alpha }$

$\therefore$ Possible values of $A$ and $B$ are
$A=\frac{\pi }{2},B=0$
and $A=\frac{\pi }{4},B=\frac{\pi }{4\sin \alpha }$