Question

If $x$ satisfies the equation $x^2{\int }_0^1\frac{dt}{t^2+2t\cos \alpha +1}-x{\int }_{-3}^3\frac{t^3\sin 2t}{t^2+1}dt-2=0,(0<\alpha <\pi )$, then the value of $x$ is?

• A. $2\sqrt{\left(\frac{\sin \alpha }{\alpha }\right)}$
• B. $-2\sqrt{\left(\frac{\sin \alpha }{\alpha }\right)}$
• C. $4\sqrt{\left(\frac{\sin \alpha }{\alpha }\right)}$
• D. None of these

Answer 1

Answer: (A), (B)

$2\sqrt{\left(\frac{\sin \alpha }{\alpha }\right)}$
$-2\sqrt{\left(\frac{\sin \alpha }{\alpha }\right)}$

Let,

$I_1={\int }_0^1\frac{dt}{t^2+2t\cos \alpha +1}{I}_2={\int }_{-3}^3\frac{t^2\sin 2t}{t^2+1}dt$

Now,

$I_1={\int }_0^1\frac{dt}{t^2+2t\cos \alpha +1}={\int }_0^1\frac{dt}{t^2+2t\cos \alpha +{\cos }^2\alpha +1-{\cos }^2\alpha }={\int }_0^1\frac{dt}{(t+\cos \alpha {)}^2+{\sin }^2\alpha }=\frac{1}{\sin \alpha }{\left[{\tan }^{-1}\frac{t+\cos \alpha }{\sin \alpha }\right]}_0^1[\text{Using}\int \frac{1}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}]=\frac{1}{\sin \alpha }\left[{\tan }^{-1}\frac{1+\cos \alpha }{\sin \alpha }\right]-\frac{1}{\sin \alpha }\left[{\tan }^{-1}\frac{\cos \alpha }{\sin \alpha }\right]=\frac{1}{\sin \alpha }\left[{\tan }^{-1}\frac{2{\cos }^2\frac{\alpha }{2}}{2\sin \frac{\alpha }{2}\cdot \cos \frac{\alpha }{2}}\right]-\frac{1}{\sin \alpha }\left[{\tan }^{-1}\frac{\cos \alpha }{\sin \alpha }\right]=\frac{1}{\sin \alpha }\left[{\tan }^{-1}\cot \frac{\alpha }{2}\right]-\frac{1}{\sin \alpha }\left[{\tan }^{-1}\cot \alpha \right]$

Given that, $0<\alpha <\pi$

$\therefore \cot \frac{\alpha }{2}=\tan (\frac{\pi }{2}-\frac{\alpha }{2})\text{and}\cot \alpha =\tan (\frac{\pi }{2}-\alpha )$

$\therefore I_1=\frac{1}{\sin \alpha }\left[{\tan }^{-1}\tan (\frac{\pi }{2}-\frac{\alpha }{2})\right]-\frac{1}{\sin \alpha }\left[{\tan }^{-1}\tan (\frac{\pi }{2}-\alpha )\right]=\frac{1}{\sin \alpha }[(\frac{\pi }{2}-\frac{\alpha }{2})-(\frac{\pi }{2}-\alpha )]=\frac{1}{\sin \alpha }\left[\frac{\alpha }{2}\right]=\frac{\alpha }{2\sin \alpha }$

Now,

$I_2={\int }_{-3}^3\frac{t^2\sin 2t}{t^2+1}dt=0[\because \frac{t^2\sin 2t}{t^2+1}\text{is an odd function}]$

Thus, the given equation reduces to

$x^2\cdot \frac{\alpha }{2\sin \alpha }-2=0\Rightarrow x^2\cdot \frac{\alpha }{2\sin \alpha }=2\Rightarrow x^2=\frac{4\sin \alpha }{\alpha }\Rightarrow x=\pm 2\sqrt{\frac{\sin \alpha }{\alpha }}$

Hence, options (A) and (B) are correct.