Answer
$\displaystyle \frac{\pi}{4}-\frac{2}{3}$
Work Step by Step
From the Strategy for Evaluating $\displaystyle \int\tan^{m}x\sec^{n}xdx$, case (a)
we use $\sec^{2}x=1+\tan^{2}x$
$\displaystyle \int_{0}^{\pi/4}\tan^{4}tdt=\int_{0}^{\pi/4}\tan^{2}t\left(\sec^{2}t-1\right)dt$
$=\displaystyle \int_{0}^{\pi/4}\tan^{2}t\sec^{2}tdt-\int_{0}^{\pi/4}\tan^{2}tdt$
... For the first integral, $\left[\begin{array}{lll}
u=\tan t & \text{bounds:} & 0, 1\\
du=\sec^{2}tdt & &
\end{array}\right],$
for the second, apply the identity
$=\displaystyle \int_{0}^{1}u^{2}du-\int_{0}^{\pi/4}\left(\sec^{2}t-1\right)dt$
$=\displaystyle \left[\frac{1}{3}u^{3}\right]_{0}^{1}-\left[\tan t-t\right]_{0}^{\pi/4}$
$=\displaystyle \frac{1}{3}-\left[\left(1-\frac{\pi}{4}\right)-0\right]$
$=\displaystyle \frac{\pi}{4}-\frac{2}{3}$
