Question

$\int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx$ is equal to

• A. $\log \left({\sin }^{-1}x\right)+c$
• B. $\frac{1}{2}{\left({\sin }^{-1}x\right)}^2+c$
• C. $\log \left(\sqrt{1-x^2}\right)+c$
• D. $\sin \left({\cos }^{-1}x\right)+c$
  where $c$ is an arbitrary constant

Answer 1

Answer: (B)

$\frac{1}{2}{\left({\sin }^{-1}x\right)}^2+c$

Let $I=\int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx$
Put ${\sin }^{-1}x=t$
$\Rightarrow \frac{1}{\sqrt{1-x^2}}dx=dt$
$\therefore I=\int tdt=\frac{t^2}{2}+c$
$=\frac{{\left({\sin }^{-1}x\right)}^2}{2}+c$