Question:
$\sec ^{2}(7-4 x)$
Solution:
Let $7-4 x=t$
$\therefore-4 d x=d t$
$\therefore \int \sec ^{2}(7-4 x) d x=\frac{-1}{4} \int \sec ^{2} t d t$
$=\frac{-1}{4}(\tan t)+\mathrm{C}$
$=\frac{-1}{4} \tan (7-4 x)+\mathrm{C}$
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