Question

The value of integral $\int \frac{\tan \sqrt{x}{\sec }^2\sqrt{x}}{\sqrt{x}}dx$ is
(where $C$ is constant of integration)

• A. $\left(\tan \sqrt{x}\right)+C$
• B. $(\tan \sqrt{x}{)}^2+C$
• C. $(\tan \sqrt{x}{)}^3+C$
• D. $(\tan \sqrt{x}{)}^4+C$

Answer 1

The correct option is B $(\tan \sqrt{x}{)}^2+C$

Put $z=\tan \sqrt{x}$
$dz={\sec }^2\sqrt{x}\cdot \frac{1}{2\sqrt{x}}dx$
$\Rightarrow I=\int 2zdz=z^2+C$
$=(\tan \sqrt{x}{)}^2+C$