Answer
$\int_{0}^{\pi/4} (1+tan~t)^3~sec^2~t~dt = \frac{15}{4}$
Work Step by Step
$\int_{0}^{\pi/4} (1+tan~t)^3~sec^2~t~dt$
Let $u = tan~t$
$\frac{du}{dt} = sec^2~t$
$dt = \frac{du}{sec^2~t}$
When $t = 0,~~$ then $~~u = 0$
When $t = \frac{\pi}{4},~~$ then $~~u = 1$
$\int_{0}^{1} (1+u)^3~sec^2~t~\frac{du}{sec^2~t}$
$= \int_{0}^{1} (1+u)^3~du$
$= \frac{1}{4}(1+u)^4~\Big\vert_{0}^{1}$
$= \frac{1}{4}(1+1)^4- \frac{1}{4}(1+0)^4$
$= \frac{1}{4}(16)- \frac{1}{4}(1)$
$= 4- \frac{1}{4}$
$= \frac{15}{4}$