Q. If $f(x) = \frac{\sin^{- 1}x}{\sqrt{1 - x^{2}}}$ and $g(x) = e^{\sin^{- 1}x}$, then $\int_{}^{}{f(x)g(x)dx}$ is equal to
A) $e^{\sin^{- 1}x}\left( \sin^{- 1}{x - 1} \right) + c$
B) $e^{\sin^{- 1}x} + c$
C) $e^{{{(sin}^{- 1}{x)\ }}^{2}} + c$
D) $e^{{2sin}^{- 1}x} + c$
Explanation
(a)
$\int_{}^{}{f(x)g(x)dx} = \int_{}^{}{\frac{\sin^{- 1}x}{\sqrt{1 - x^{2}}}\ e^{\sin^{- 1}x}\ }dx$
Put $\sin^{- 1}{x = t}\ \ \ \ \Rightarrow \ \ \ \frac{1}{\sqrt{1 - x^{2}}}dx = dt\$
$\therefore\ \int_{}^{}{f(x)}g(x)dx = \int_{}^{}{te^{t}}dt = te^{t} - e^{t} + c\$
$= e^{\sin^{- 1}x}\left( \sin^{- 1}{x - 1} \right) + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.