Question

lf ${\int }_0^1\frac{{\tan }^{-1}x}{x}dx=k{\int }_0^{\pi /2}\frac{x}{\sin x}dx$, then the value of $k$ is

• A. $1$
• B. $\frac{1}{4}$
• C. $4$
• D. $2$

Answer 1

Answer: (D)

$2$

Let $I={\int }_0^{\frac{\pi }{2}}\frac{x}{sinx}dx$
Substitute $tan\left(\frac{x}{2}\right)=t\Rightarrow \frac{1}{2}{sec}^2\left(\frac{x}{2}\right)dx=dt$
gives $sinx=\frac{2t}{1+t^2},dx=\frac{2dt}{1+t^2}$
$I={\int }_0^1\frac{2{tan}^{-1}t}{\frac{2t}{1+t^2}}\frac{2dt}{1+t^2}=2{\int }_0^1\frac{{tan}^{-1}t}{t}dt=2{\int }_0^1\frac{{tan}^{-1}x}{x}dx\Rightarrow k=2$