Question

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

• A. $\sin \alpha$
• B. $\alpha \sin \alpha$
• C. $\frac{3\alpha -\pi }{2\sin \alpha }$
• D. $\frac{\alpha }{2}\sin \alpha$

Answer 1

The correct option is C $\frac{3\alpha -\pi }{2\sin \alpha }$

${\int }_0^1\frac{dx}{x^2+2x\cos \alpha +1}$

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

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

As we know that,

$\int \frac{dx}{a^2+x^2}=\frac{1}{a}\left({\tan }^{-1}\frac{x}{a}\right)$

$\therefore {\int }_0^1\frac{dx}{{(x+\cos \alpha )}^2+{\sin }^2\alpha }={\left[\frac{1}{\sin \alpha }\left({\tan }^{-1}\left(\frac{x+\cos \alpha }{\sin \alpha }\right)\right)\right]}_0^1$

$\Rightarrow =\frac{1}{\sin \alpha }[{\tan }^{-1}\left(\frac{1+\cos \alpha }{\sin \alpha }\right)-{\tan }^{-1}\left(\frac{0+\cos \alpha }{\sin \alpha }\right)]$

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

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

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

$\Rightarrow =\frac{1}{\sin \alpha }(\frac{3\alpha }{2}-\frac{\pi }{2})$

$\Rightarrow =\frac{3\alpha -\pi }{2\sin \alpha }$