Question

The value of ${\int }_0^{1/\sqrt{2}}\frac{{\sin }^{-1}x}{{(1-x^2)}^{3/2}}dx$ is

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

Answer 1

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

Let $l={\int }_0^{1/\sqrt{2}}\frac{{\sin }^{-1}x}{{(1-x^2)}^{3/2}}dx$

When $x=0$, then $\sin t=0=\sin 0\Rightarrow t=0$

When $x=\frac{1}{\sqrt{2}}$, then $\sin t=\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}$

$\Rightarrow t=\frac{\pi }{4}$

Put $x=\sin tdx=\cos tdt$

$\therefore l={\int }_0^{\pi /4}\frac{t}{{\cos }^{3}t}\cdot \cos tdt={\int }_0^{\pi /4}t{\sec }^{2}tdt$

$={[t\cdot \tan t]}_0^{\pi /4}-{\int }_0^{\pi /4}\tan tdt$

$=\frac{\pi }{4}-{\left[\log \sec t\right]}_0^{\pi /4}$

$=\frac{\pi }{4}-\log \sqrt{2}-0=\frac{\pi }{4}-\log 2^{1/2}=\frac{\pi }{4}-\frac{1}{2}\log 2$