Question

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

Answer 1

$I_1={\int }_0^1\frac{{\tan }^{-1}xdx}{x}$

Let $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$

When $x=0\Rightarrow \theta =0$

When $x=1\Rightarrow \theta =\frac{\pi }{4}$

$={\int }_0^{\frac{\pi }{4}}\frac{\theta {\sec }^2\theta d\theta }{\tan \theta }$

$={\int }_0^{\frac{\pi }{4}}\frac{\theta \frac{1}{{\cos }^2\theta }d\theta }{\frac{\sin \theta }{\cos \theta }}$

$={\int }_0^{\frac{\pi }{4}}\frac{2\theta d\theta }{\sin 2\theta },$

Let $2\theta =t\Rightarrow 2d\theta =dt$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{tdt}{\sin t}=\frac{1}{2}{I}_2$

$\Rightarrow \frac{I_1}{I_2}=\frac{1}{2}$