Question

The value of the integral $\underset{0}{\overset{\frac{1}{\sqrt{2}}}{\int }}\frac{{\sin }^{-1}xdx}{(1-x^2{)}^{\frac{3}{2}}}$

• A. $\frac{\pi }{2}+\frac{1}{2}\log 2$
• B. $\pi -\frac{1}{2}\log 2$
• C. $\frac{\pi }{2}-\log 2$
• D. $\frac{\pi }{4}-\frac{1}{2}\log 2$

Answer 1

The correct option is D $\frac{\pi }{4}-\frac{1}{2}\log 2$

$I=\underset{0}{\overset{\frac{1}{\sqrt{2}}}{\int }}\frac{{\sin }^{-1}xdx}{(1-x^2{)}^{\frac{3}{2}}}$
Let $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$
$I=\underset{0}{\overset{\pi /4}{\int }}\frac{\theta \cos \theta d\theta }{(1-{\sin }^2\theta {)}^{\frac{3}{2}}}$
$I=\underset{0}{\overset{\pi /4}{\int }}\theta {\sec }^2\theta d\theta$
$=[\theta \tan \theta {]}_0^{\pi /4}-\underset{0}{\overset{\frac{\pi }{4}}{\int }}1\cdot \tan \theta d\theta =\frac{\pi }{4}-[\log \sec \theta {]}_0^{\pi /4}$
$=\frac{\pi }{4}-[\log \sec \frac{\pi }{4}-\log \sec 0]$
$=\frac{\pi }{4}-[\log \sqrt{2}-\log 1]$
$=\frac{\pi }{4}-\frac{1}{2}\log 2$