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 $\frac{\alpha }{2\sin \alpha }$

${\int }_0^1\frac{dx}{x^2+2xcos\alpha +1}={\int }_0^1\frac{dx}{x^2+2xcos\alpha +co{s}^2\alpha +si{n}^2\alpha }$

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

$=\frac{1}{sin\alpha }{\left[ta{n}^{-1}\left(\frac{x+cos\alpha }{sin\alpha }\right)\right]}_0^1$

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

$=\frac{1}{sin\alpha }[ta{n}^{-1}\left(\frac{2co{s}^2\alpha /2}{2sin\alpha /2cos\alpha /2}\right)-ta{n}^{-1}\left(cot\alpha \right)]$

$=\frac{1}{sin\alpha }[ta{n}^{-1}(cot+\alpha /2)-ta{n}^{-1}\left(cot\alpha \right)]$

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

$=\frac{1}{sin\alpha }[\pi /2-\alpha /2-\pi /2+\alpha ]$

$=\frac{\alpha }{2sin\alpha }$