Question

If $I_1={\int }_0^{\pi /2}\frac{x}{\sin x}dx$ and $I_2={\int }_0^{\pi /2}\frac{{\tan }^{-1}x}{x}dx,$ then $\frac{I_1}{I_2}=$

• A. $\frac{1}{2}$
• B. 1
• C. 2
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C 2

Substitute $x=\tan \theta \therefore I_2={\int }_0^{\pi /4}\frac{\theta }{\tan \theta }.{\sec }^2\theta d\theta$ or $I_2={\int }_0^{\pi /4}\frac{2\theta }{2\sin \theta \cos \theta }d\theta ={\int }_0^{\pi /4}\frac{2\theta }{\sin 2\theta }d\theta$
Now substitute $2\theta =t\therefore I_2=\frac{1}{2}{\int }_0^{\pi /2}\frac{t}{\sin t}dt=\frac{1}{2}{I}_1$ $\therefore \frac{I_1}{I_2}=2\Rightarrow \left(C\right)$