Q. $\int_{}^{}{\frac{\sin^{- 1}x}{\sqrt{1 - x^{2}}}\ dx}$ is equal to
A) $\log{(\sin^{- 1}{x)}} + c$
B) $\frac{1}{2}\ \left( \sin^{- 1}x \right)^{2} + c$
C) $\log\left( \sqrt{1 - x^{2}} \right) + c$
D) $\sin{(\cos^{- 1}x}) + c$
Explanation
(b)
Let $I = \int_{}^{}{\frac{\sin^{- 1}x}{\sqrt{1 - x^{2}}}\ dx}$
Put $\sin^{- 1}x = t$
$\Rightarrow \ \frac{1}{\sqrt{1 - x^{2}}}dx = dt$
$\therefore\ I = \int_{}^{}tdt = \frac{t^{2}}{2} + c$
$= \frac{\left( \sin^{- 1}x \right)^{2}}{2} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.