Question

Solve: ${\int }_0^1\frac{{\tan }^{-1}x}{x}dx=.......$

• A. ${\int }_0^{\pi /4}\frac{\sin x}{x}dx$
• B. ${\int }_0^{\pi /2}\frac{x}{\sin x}dx$
• C. $\frac{1}{2}{\int }_0^{\pi /2}\frac{x}{\sin x}dx$
• D. $\frac{1}{2}{\int }_0^{\pi /4}\frac{x}{\sin x}dx$

Answer 1

The correct option is C $\frac{1}{2}{\int }_0^{\pi /2}\frac{x}{\sin x}dx$

Consider, $I={\int }_0^1\frac{{\text{tan}}^{-1}x}{x}dx$

put $x=\tan t⟹dx={\text{sec}}^{2}tdt$

when $x=1,t=\frac{\pi }{4}$ and when $x=0,t=0$

${\int }_0^{\frac{\pi }{4}}\frac{t}{\tan t}{\text{sec}}^{2}tdt=\int \frac{t}{\sin t\cos t}dt=\int \frac{2t}{\sin 2t}dt$

put $x=2t⟹dx=2dt$

$t=\frac{\pi }{4},x=\frac{\pi }{2}$ and $t=0,x=0$

${\int }_0^{\pi /2}\frac{x}{\sin x}\times \frac{dx}{2}=\frac{1}{2}{\int }_0^{\pi /2}\frac{x}{\sin x}dx$