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