Prove

Let $7-4 x=t$

$\therefore-4 d x=d t$

$\therefore \int \sec ^{2}(7-4 x) d x=\frac{-1}{4} \int \sec ^{2} t d t$

$=\frac{-1}{4}(\tan t)+\mathrm{C}$

$=\frac{-1}{4} \tan (7-4 x)+\mathrm{C}$