Question

The value of the integral $\underset{2/\pi }{\overset{\pi /2}{\int }}\frac{{\tan }^{-1}x}{x}dx$ is

• A. $\frac{\pi }{4}\ln \left(\frac{\pi }{2}\right)$
• B. $\frac{\pi }{2}\ln \left(\frac{\pi }{2}\right)$
• C. $\pi \ln \left(\frac{\pi }{2}\right)$
• D. $\frac{1}{2}\ln \left(\frac{\pi }{2}\right)$

Answer 1

The correct option is B $\frac{\pi }{2}\ln \left(\frac{\pi }{2}\right)$

$\text{Let}I=\underset{2/\pi }{\overset{\pi /2}{\int }}\frac{{\tan }^{-1}x}{x}dx\cdots \left(1\right)$
Put $x=\frac{1}{t}\Rightarrow dx=\frac{-1}{t^2}dt$
$I=\underset{\pi /2}{\overset{2/\pi }{\int }}t{\cot }^{-1}t\left(\frac{-1}{t^2}dt\right)$
$=\underset{\pi /2}{\overset{2/\pi }{\int }}\frac{-{\cot }^{-1}t}{t}dt$
$\Rightarrow I=\underset{2/\pi }{\overset{\pi /2}{\int }}\frac{{\cot }^{-1}t}{t}dt=\underset{2/\pi }{\overset{\pi /2}{\int }}\frac{{\cot }^{-1}x}{x}dx\cdots \left(2\right)$

Adding $\left(1\right)$ and $\left(2\right),$ we get
$2I=\underset{2/\pi }{\overset{\pi /2}{\int }}\frac{\pi /2}{x}dx$
$\Rightarrow I=\frac{\pi }{4}\ln \frac{{\pi }^2}{4}=\frac{\pi }{2}\ln \left(\frac{\pi }{2}\right)$