Answer
$$\displaystyle{\int_0^{\frac{\pi}{4}}{\tan ^3\theta \sec ^2\theta }\ d\theta=\frac{1}{4}}$$
Work Step by Step
$I=\displaystyle\int_0^{\frac{\pi}{4}}{\tan ^3\theta \sec ^2\theta d\theta}$
Consider:
$\displaystyle\frac{d}{dx}\tan ^4x=4\tan ^3x\sec ^2x$
$\displaystyle{I=\int_0^{\frac{\pi}{4}}{\frac{4}{4}\tan ^3\theta \sec ^2\theta d\theta}\\
I=\frac{1}{4}\int_0^{\frac{\pi}{4}}{4\tan ^3\theta \sec ^2\theta d\theta}\\
I=\frac{1}{4}\left[ \tan ^4\theta \right] _{0}^{\frac{\pi}{4}}\\
I=\frac{1}{4}}$
