Question

Integrate $\frac{{\sec }^2({\sin }^{-1}x)}{\sqrt{1-x^2}}$

• A. $\sin ({\tan }^{-1}x)+c$
• B. $\tan ({\sec }^{-1}x)+c$
• C. $\tan ({\sin }^{-1}x)+c$
• D. $-\tan (co{s}^{-1}x)+c$

Answer 1

Answer: (C)

Let $l=\int \frac{{\sec }^2({\sin }^{-1}x)}{\sqrt{1-x^2}}dx$ Again, let ${\sin }^{-1}x=t$
$\Rightarrow$ $\frac{dt}{dx}=\frac{1}{\sqrt{1-x^2}}$
$\Rightarrow$ $dt=\frac{1}{\sqrt{1-x^2}}dx$
$\therefore$ $l=\int {\sec }^{2}tdt$
$=\tan t+C$
$=\tan ({\sin }^{-1}x)+C$