Question

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

The correct option is D $\pm 2\sqrt{\frac{\sin \alpha }{\alpha }}$

$\left[\underset{0}{\overset{1}{\int }}\frac{dt}{t^2+2t\cos \alpha +1}\right]x^2-\left[\underset{-3}{\overset{3}{\int }}\frac{t^2\sin 2t}{t^2+1}dt\right]x-2=0$
We know,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=0\text{if}f\left(x\right)+f(-x)=0$
Let,
$I=\underset{0}{\overset{1}{\int }}\frac{dt}{t^2+2t\cos \alpha +1}\Rightarrow I=\underset{0}{\overset{1}{\int }}\frac{dt}{(t+\cos \alpha {)}^2+{\sin }^2\alpha }\Rightarrow I=\frac{1}{\sin \alpha }{\left[{\tan }^{-1}\left(\frac{t+\cos \alpha }{\sin \alpha }\right)\right]}_0^1\Rightarrow I=\frac{1}{\sin \alpha }[{\tan }^{-1}\cot \frac{\alpha }{2}-{\tan }^{-1}\cot \alpha ]\Rightarrow I=\frac{1}{\sin \alpha }[{\tan }^{-1}\tan (\frac{\pi }{2}-\frac{\alpha }{2})-{\tan }^{-1}\tan (\frac{\pi }{2}-\alpha )]\Rightarrow I=\frac{1}{\sin \alpha }[(\frac{\pi }{2}-\frac{\alpha }{2})-(\frac{\pi }{2}-\alpha )]\therefore I=\frac{\alpha }{2\sin \alpha }$

From the equation, we get
$I\cdot x^2-2=0\Rightarrow x^2=\frac{4\sin \alpha }{\alpha }\therefore x=\pm 2\sqrt{\frac{\sin \alpha }{\alpha }}$