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