Answer
$$ \frac{\pi }{4}-\frac{2}{3}$$
Work Step by Step
Given
$$\int_{0}^{\pi/4}\tan^4tdt $$
Since
$$1+\tan^2t=\sec^2 t $$
Then
\begin{align*}
\int_{0}^{\pi/4}\tan^4tdt &=\int_{0}^{\pi/4}\tan^2t\tan^2tdt \\
&=\int_{0}^{\pi/4}\tan^2t(\sec^2t-1)dt \\
&=\int_{0}^{\pi/4}(\tan^2t\sec^2t-\tan^2t)dt \\
&=\int_{0}^{\pi/4}(\tan^2t\sec^2t-\sec^2t+1)dt \\
&=\frac{1}{3}\tan^3t-\tan t+t\bigg|_{0}^{\pi/4}\\
&= \frac{\pi }{4}-\frac{2}{3}
\end{align*}
