Q. The value of $\int_{}^{}\frac{d(\sin{x)}}{\sqrt{1 - \sin^{2}{x\ }}}$ is equal to
A) $x + c$
B) $3x + c$
C) $x^{2} + c$
D) None of these
Explanation
(a)
Let $\sin{x = \mathcal{z\ \Rightarrow}d\ (\sin{x) = d\mathcal{z}}}$
$\therefore\ \int_{}^{}{\frac{d\mathcal{z}}{\sqrt{1 - \mathcal{z}^{2}}} = \sin^{- 1}{\mathcal{z +}c}}$
$= \sin^{- 1}{(\sin{x) + c = x + c}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.