$\huge \int_{ 0 } ^ { \pi/4} 3.5 \sec^2 (x) \tan^{4} (x) \, \d x $ $\huge \begin{align} \let u &= \tan x \\ \let \d u &= \sec^2 (x) \d x \\ \int_{ 0 } ^ { \pi/4} 3.5 \sec^2 (x) \tan^{4} (x) \, \d x &= \frac{7}{2} \int_{\tan 0}^{\tan \frac{\pi}{4} } u^4 \d u \\ &= \frac{7}{2} \pa{ \frac{1}{5} - 0 } \\ &= \frac{7}{10} \end{align} $