A) \[\frac{\pi }{4}+\frac{1}{2}\log 2\]
B) \[\frac{\pi }{4}-\frac{1}{2}\log 2\]
C) \[\frac{\pi }{2}+\log 2\]
D) \[\frac{\pi }{2}-\log 2\]
Correct Answer: B
Solution :
\[I=\int_{0}^{1/\sqrt{2}}{\frac{{{\sin }^{-1}}x}{{{(1-{{x}^{2}})}^{3/2}}}}dx\] Put \[{{\sin }^{-1}}x=t\Rightarrow \frac{1}{\sqrt{1-{{x}^{2}}}}dx=dt\]and \[x=\sin t\] Also \[t=0\]to \[\frac{\pi }{4}\]as \[x=0\]to \[\frac{1}{\sqrt{2}}\] \[\Rightarrow I=\int_{0}^{\pi /4}{t.{{\sec }^{2}}t\,dt=\frac{\pi }{4}-\frac{1}{2}\log 2}\].
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