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. $\frac{\alpha }{2\sin \alpha }$
• D. $\frac{\alpha }{2}\sin \alpha$

Answer 1

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

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

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

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

$⟹\frac{1}{\sin \alpha }[{\tan }^{-1}\frac{x+\cos x}{\sin x}{]}_0^1$ $[\because \int \frac{1}{x^2+a^2}=\frac{1}{a}ta{n}^{-1}(\frac{x}{a}\left)\right]$

$⟹\frac{1}{\sin \alpha }[{\tan }^{-1}\frac{1+\cos \alpha }{\sin \alpha }-{\tan }^{-1}\frac{\cos \alpha }{\sin \alpha }]$

$⟹\frac{1}{\sin \alpha }[{\tan }^{-1}\frac{\frac{1+\cos \alpha }{\sin \alpha }-\frac{\cos \alpha }{\sin \alpha }}{1+\frac{\cos \alpha +{\cos }^2\alpha }{{\sin }^2\alpha }}]$

$⟹\frac{1}{\sin \alpha }{\tan }^{-1}\frac{\sin \alpha }{1+\cos \alpha }$

$⟹\frac{1}{\sin \alpha }{\tan }^{-1}\tan \frac{\alpha }{2}$

$⟹\frac{\alpha }{2\sin \alpha }$