Question

$\int \frac{1}{{\sin }^{-1}x\sqrt{(1-x^2)}}dx.$

• A. $\log {\sin }^{-1}x.$
• B. $\log \sin x.$
• C. $\frac{1}{2}\log {\sin }^{-1}x.$
• D. $\log {\sin }^{-1}{x}^2.$

Answer 1

The correct option is A $\log {\sin }^{-1}x.$

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