Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 24

Find the indefinite integral $\displaystyle \int \frac{dt}{\cos ^2 t \sqrt{1 + \tan t}} dx$


$
\begin{equation}
\begin{aligned}

\int \frac{dt}{\cos ^2 t \sqrt{1 + \tan t}} =& \int \frac{1}{cos ^2 t \sqrt{1 + \tan t}} dt
\\
\\
\int \frac{dt}{\cos ^2 t \sqrt{1 + \tan t}} =& \int \frac{\sec ^2 t}{\sqrt{1 + \tan t}} dt

\end{aligned}
\end{equation}
$



If we let $\displaystyle u = 1 + \tan t$, then $\displaystyle du = \sec^2 t dt$. And


$
\begin{equation}
\begin{aligned}

\int \frac{\sec ^2 t}{\sqrt{1 + \tan t}} dt =& \int \frac{1}{\sqrt{1 + \tan t}} \sec ^2 t dt
\\
\\
\int \frac{\sec ^2 t}{\sqrt{1 + \tan t}} dt =& \int \frac{1}{\sqrt{u}} du
\\
\\
\int \frac{\sec ^2 t}{\sqrt{1 + \tan t}} dt =& \int u^{\frac{-1}{2}} du
\\
\\
\int \frac{\sec ^2 t}{\sqrt{1 + \tan t}} dt =& \frac{u^{\frac{-1}{2} + 1} }{\displaystyle \frac{-1}{2} + 1} + C
\\
\\
\int \frac{\sec ^2 t}{\sqrt{1 + \tan t}} dt =& \frac{u^{\frac{1}{2}}}{\displaystyle \frac{1}{2}} + C
\\
\\
\int \frac{\sec ^2 t}{\sqrt{1 + \tan t}} dt =& 2u^{\frac{1}{2}} + C
\\
\\
\int \frac{\sec ^2 t}{\sqrt{1 + \tan t}} dt =& 2 (1 + \tan t)^{\frac{1}{2}} + C

\end{aligned}
\end{equation}
$