Question

Evaluate the integral

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

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

Answer 1

The correct option is C $\frac{\alpha }{\left(2\sin \alpha \right)}$

${\int }_0^1\frac{dx}{x^2+xco{s}_{\alpha }+1}$
${\int }_0^1\frac{dx}{(x+cos\alpha {)}^2}+si{n}^2\alpha$
${(\frac{1}{sin\alpha }\left[ta{n}^{-1}\left(\frac{x+cos\alpha }{sin\alpha }\right)\right]+c)}_0^1$
$=\frac{1}{sin\alpha }\left[ta{n}^{-1}\left(\frac{1+cos\alpha }{sin\alpha }\right)\right]-\frac{1}{sin\alpha }ta{n}^{-1}\left(\frac{cos\alpha }{sin\alpha }\right)$
$=\frac{1}{sin\alpha }\left[ta{n}^{-1}\right(cot\frac{\alpha }{2})-ta{n}^{-1}(cot\alpha \left)\right]$
$=\frac{1}{sin\alpha }(\alpha -\frac{\alpha }{2})=\frac{\alpha }{2}sin\alpha$