Question

Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]

 

Answer 1

\[\text{ Let I }= \int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]
\[\text{ Let }\sqrt{x} = t\]
\[ \Rightarrow \frac{dx}{2\sqrt{x}} = dt\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2\text{  dt}\]
\[\text{ Putting}\ \sqrt{x} = t  \text{ and} \frac{dx}{\sqrt{x}} = \text{ 2 dt }\]
\[ \therefore I = 2\int \sec^2 + dt\]
\[ = 2 \tan t + C\]
\[ = 2 \tan \left( \sqrt{x} \right) + C \left( \because t = \sqrt{x} \right)\]