Question

The value of the definite integral ${\int }_0^1\frac{dx}{x^2+2x\cos \alpha +1}$ for $0<\alpha <\pi$ is equal to

• A. $\sin \alpha$
• B. $ta{n}^{-1}\left(\sin \alpha \right)$
• C. $\alpha \sin \alpha$
• D. $\frac{\alpha }{2}(\sin \alpha {)}^{-1}$

Answer 1

The correct option is D $\frac{\alpha }{2}(\sin \alpha {)}^{-1}$

${\int }_0^1\frac{dx}{x^2+2x\cos \alpha +1},0<\alpha <\pi$

$={\int }_0^1\frac{dx}{x^2+2x.\cos \alpha +(\cos \alpha {)}^2+1-(\cos \alpha {)}^2}$

$={\int }_0^1\frac{dx}{(x+\cos \alpha {)}^2+(1-{\cos }^2\alpha )}$

$={\int }_0^1\frac{dx}{(x+\cos \alpha {)}^2+(\sin \alpha {)}^2}$

Let, $x+\cos \alpha =z\Rightarrow dx=dx$

$={\int }_{cos\alpha }^{1+\cos \alpha }\frac{dz}{(z{)}^2+\sin \alpha {)}^2}$

$=\frac{1}{\sin \alpha }{\left[{\tan }^{-1}\left(\frac{z}{\sin \alpha }\right)\right]}_{\cos \alpha }^{1+\cos \alpha }$

$=\frac{1}{\sin \alpha }[{\tan }^{-1}\left(\frac{1+\cos \alpha }{\sin \alpha }\right)-{\tan }^{-1}\left(\frac{\cos \alpha }{\sin \alpha }\right)]$

$=\frac{1}{\sin \alpha }[{\tan }^{-1}\left(\frac{2{\cos }^2\frac{\alpha }{2}}{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}\right)-{\tan }^{-1}\left(\cot \alpha \right)]$

$=\frac{1}{\sin \alpha }[{\tan }^{-1}\left(cot\frac{\alpha }{2}\right)-{\tan }^{-1}\left(\cot \alpha \right)]$

$=\frac{1}{\sin \alpha }[(\frac{\pi }{2}-\frac{\alpha }{2})-(\frac{\pi }{2}-\alpha )]$

$=\frac{1}{\sin \alpha }[-\frac{\alpha }{2}+\alpha ]=\frac{\alpha }{2}(\sin \alpha {)}^{-1}$