Answer
$$ \frac{15}{4}$$
Work Step by Step
Given
$$\int_{0}^{\pi/4} (1+\tan t )^3\sec^2 tdt$$
Let $x=1+\tan t\ \Rightarrow \ dx=\sec^2 tdt $, at $t=0\to x=1$, at $t=\pi/4\to x= 2$, then
\begin{align*}
\int_{0}^{\pi/4} (1+\tan t )^3\sec^2 tdt&=\int_{1}^{2}u^3 du\\
&=\frac{1}{4}u^4\bigg|_{1}^{2}\\
&= \frac{15}{4}
\end{align*}
