Question

If ${\int }_0^{\alpha }\frac{dx}{1-\cos \alpha \cos x}=\frac{A}{\sin \alpha }+B(a\ne 0)$
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)

$A=\frac{\pi }{2},B=0$
$A=\frac{\pi }{4},B=\frac{\pi }{4\sin \alpha }$

Let $I={\int }_0^{\alpha }\frac{dx}{1-\cos \alpha \cos \alpha }$
$={\int }_0^{\alpha }\frac{dx}{({\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2})-\cos \alpha ({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})}$
$={\int }_0^{\alpha }\frac{dx}{(1-\cos \alpha ){\cos }^2\frac{x}{2}+(1+\cos \alpha ){\sin }^2\frac{x}{2}}$
$={\int }_0^{\alpha }\frac{dx}{2{\sin }^2\frac{\alpha }{2}{\cos }^2\frac{x}{2}+2{\cos }^2\frac{\alpha }{2}{\sin }^2\frac{x}{2}}$
$=\frac{1}{2}{\int }_0^{\alpha }\frac{{\sec }^2\frac{\alpha }{2}{\sec }^2\frac{x}{2}}{{\tan }^2\frac{\alpha }{2}+{\tan }^2\frac{x}{2}}dx$
put $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
$\therefore I={\int }_0^{\tan \frac{\alpha }{2}}\frac{{\sec }^2\frac{\alpha }{2}}{t^2+{\tan }^2\frac{\alpha }{2}}dt$
$={\sec }^2\frac{\alpha }{2}\cot \frac{\alpha }{2}{\left[{\tan }^{-1}\left(\frac{t}{\tan \frac{\alpha }{2}}\right)\right]}_0^{\tan \frac{\alpha }{2}}$
$=\frac{2}{\sin \alpha }.\frac{\pi }{4}=\frac{\pi }{2\sin \alpha }$
$A=\frac{\pi }{2},B=0$ or $A=\frac{\pi }{4},B=\frac{\pi }{4\sin \alpha }$