Answer
$$-\frac{1}{4}+\frac{1}{2}\ln \left(2\right)$$
Work Step by Step
\begin{align*}
\int_{0}^{\pi / 4} \tan^5xdx&=\int_{0}^{\pi / 4} \tan^3x \tan^2x dx\\
&=\int_{0}^{\pi / 4} \tan^3x ( \sec^2x-1) dx\\
&=\int_{0}^{\pi / 4} \tan^3x \sec^2x-\int_{0}^{\pi / 4} \tan^3x dx\\
&=\int_{0}^{\pi / 4} \tan^3x \sec^2x-\int_{0}^{\pi / 4} (\sec^2 x-1)\tan x dx\\
&= \frac{1}{4}\tan^4 x-\frac{1}{2}\tan^2 x +\ln \sec x\bigg|_{0}^{\pi/4}\\
&=-\frac{1}{4}+\frac{1}{2}\ln \left(2\right)
\end{align*}
