Answer
$\displaystyle \frac{\pi}{4}-\frac{1}{2}$
Work Step by Step
$\displaystyle \int_{0}^{\pi/4}\frac{x\sin x}{\cos^{3}x}dx=\int_{0}^{\pi/4}\frac{x\sin x}{\cos x}\cdot\sec^{2}xdx=\int_{0}^{\pi/4}x\cdot\tan x\cdot\sec^{2}xdx$
By parts, $\displaystyle \left[\begin{array}{ll}
u=x, & dv=\tan x\sec^{2}xdx\\
du=dx & v=\frac{1}{2}\tan^{2}x
\end{array}\right],\quad\int_{a}^{b}udv=uv|_{a}^{b}-\int_{a}^{b}vdu$
$=\displaystyle \left[\frac{x}{2}\tan^{2}x\right]_{0}^{\pi/4}-\frac{1}{2}\int_{0}^{\pi/4}\tan^{2}xdx$
$=(\displaystyle \frac{\pi}{8}\cdot 1-0)-\frac{1}{2}\int_{0}^{\pi/4}(\sec^{2}x-1)dx$
$=\displaystyle \frac{\pi}{8}-\frac{1}{2}\left[\tan^{2}x-x\right]_{0}^{\pi/4}$
$=\displaystyle \frac{\pi}{8}-\frac{1}{2}[1-\frac{\pi}{4}-(0)]$
$=\displaystyle \frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{4}$
$=\displaystyle \frac{\pi}{4}-\frac{1}{2}$
