Question

If $x$ satisfies the equation $\left(\underset{0}{\overset{1}{\int }}\frac{dt}{t^2+2t\cos \alpha +1}\right)x^2-\left(\underset{-3}{\overset{3}{\int }}\frac{t^2\sin 2t}{t^2+1}dt\right)x-2=0(0<\alpha <\pi )$, then the value $x$ is

• A. $\pm \sqrt{\frac{\alpha }{2\sin \alpha }}$
• B. $\pm \sqrt{\frac{2\sin \alpha }{\alpha }}$
• C. $\pm \sqrt{\frac{\alpha }{\sin \alpha }}$
• D. $\pm 2\sqrt{\frac{\sin \alpha }{\alpha }}$

Answer 1

Answer: (D)

$\underset{-3}{\overset{3}{\int }}\frac{t^2\sin 2t}{t^2+1}dt=0$ as the integrand is an odd function.
also $\underset{0}{\overset{1}{\int }}\frac{dt}{t^2+2t\cos \alpha +1}={\frac{1}{\sin \alpha }{\tan }^{-1}\frac{t+\cos \alpha }{\sin \alpha }|}_0^1=\frac{\alpha }{2\sin \alpha }$
Thus the given equation reduces to
$x^2\frac{\alpha }{2\sin \alpha }-2=0\Rightarrow x=\pm 2\sqrt{\frac{\sin \alpha }{\alpha }}$