Question

Evaluate the integral

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

• A. $\frac{\pi }{4}+\frac{1}{2}$ log 2
• B. $\frac{\pi }{4}-\frac{1}{2}$ log 2
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (B)

$\frac{\pi }{4}-\frac{1}{2}$ log 2

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

Substituting $x=sint\Rightarrow dx=costdt$

$I={\int }_0^{\frac{\pi }{4}}\frac{tcostdt}{cost-{sin}^{2}tcost}={\int }_0^{\frac{\pi }{4}}t{sec}^{2}tdt={\left[ttant\right]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}tantdt=\frac{\pi }{4}-0-{[-\log \left(cosx\right)]}_0^{\frac{\pi }{4}}=\frac{\pi }{4}+\log \frac{1}{\sqrt{2}}=\frac{\pi }{4}-\frac{1}{2}log2$