Question

Evaluate: $\int \frac{x}{\sqrt{(1-x^2)}{\cos }^2\sqrt{(1-x^2)}}dx$

• A. $\tan \sqrt{1-x^2}$
• B. $-\tan \sqrt{1-x^2}$
• C. $-\tan \left(1-x^2\right)$
• D. $-{\sec }^2\sqrt{1-x^2}$

Answer 1

Answer: (B)

$-\tan \sqrt{1-x^2}$

$I=\int \frac{xdx}{\sqrt{(1-x^2)}{\cos }^2\sqrt{(1-x^2)}}=\int \frac{x}{\sqrt{(1-x^2)}}{\sec }^2\sqrt{1-x^2}dx.$
Put $\sqrt{1-x^2}=t\therefore \frac{1}{2\sqrt{(1-x^2)}}=t\therefore \frac{1}{2\sqrt{(1-x^2)}}(-2x)dx=dt.$
$I=-\int {\sec }^{2}tdt=-\tan t=-\tan \sqrt{1-x^2}.$