Question

The value of definite integral $2 \int\limits_0^{\frac{1}{\sqrt{2}}} \frac{\sin ^{-1} x}{x} d x-\int\limits_0^1 \frac{\tan ^{-1} x}{x} d x$ 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
$I=\int\limits_0^{\pi / 4}(2 t) \cdot(\cot 2 t) d t$
Put $2 t=y$; we get
$I =\frac{1}{2} \int\limits_0^{\pi / 2} \underset{(I)}{y} \cdot(\underset{(II)}{\cot} y ) dy =\frac{\pi}{4} \ln 2$