Question

The value of definite integral $2\underset{0}{\overset{\frac{1}{\sqrt{2}}}{\int }}\frac{{\sin }^{-1}x}{x}dx-\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}x}{x}dx$ is equal to

• A. $\frac{\pi }{8}\ln 2$
• B. $\frac{\pi }{4}\ln 2$
• C. $\frac{\pi }{2\sqrt{2}}\ln 2$
• D. $\frac{\pi }{2}\ln 2$

Answer 1

Answer: (B)

Put ${\sin }^{-1}x=t$ in first integral and ${\tan }^{-1}x=t$ in second integral, we get
$I=\underset{0}{\overset{\pi /4}{\int }}(2t)\cdot (\cot 2t)dt$
Put $2t=y$; we get
$I=\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}\underset{(I)}{y}\cdot (\underset{(II)}{\cot }y)dy=\frac{\pi }{4}\ln 2$