Question

The value of the integral $\underset{0}{\overset{1}{\int }}\frac{dx}{x^2+2x\cos \alpha +1},\alpha \in (0,\frac{\pi }{2})$ is equal to

• A. $\sin \alpha$
• B. $\alpha \sin \alpha$
• C. $\frac{\alpha }{2\sin \alpha }$
• D. $\frac{\alpha }{2}\sin \alpha$

Answer 1

The correct option is C $\frac{\alpha }{2\sin \alpha }$

$I=\underset{0}{\overset{1}{\int }}\frac{dx}{(x+\cos \alpha {)}^2+(1-{\cos }^2\alpha )}$
$=\underset{0}{\overset{1}{\int }}\frac{dx}{(x+\cos \alpha {)}^2+{\sin }^2\alpha }$
$=\frac{1}{\sin \alpha }{\left[{\tan }^{-1}\frac{x+\cos \alpha }{\sin \alpha }\right]}_0^1$
$=\frac{1}{\sin \alpha }[{\tan }^{-1}\left(\frac{1+\cos \alpha }{\sin \alpha }\right)-{\tan }^{-1}\left(\frac{\cos \alpha }{\sin \alpha }\right)]$
$=\frac{1}{\sin \alpha }[{\tan }^{-1}\left(\cot \frac{\alpha }{2}\right)-{\tan }^{-1}(\cot \alpha \left)\right]$
$=\frac{1}{\sin \alpha }[{\tan }^{-1}\left(\tan (\frac{\pi }{2}-\frac{\alpha }{2})\right)-{\tan }^{-1}\left(\tan (\frac{\pi }{2}-\alpha )\right)]$
$=\frac{1}{\sin \alpha }[(\frac{\pi }{2}-\frac{\alpha }{2})-(\frac{\pi }{2}-\alpha )]$
$=\frac{\alpha }{2\sin \alpha }$