Question:
Evaluate: $\int \sec ^{2}(7-4 x) d x$
Solution:
let $7-4 x=t$
Differentiating on both sides we get,
$-4 d x=d t$
$d x=-\frac{1}{4} d t$
substituting it in $\int \sec ^{2}(7-4 x) d x$ we get,
$=\int-\frac{1}{4} \sec ^{2} t d t$
$=\tan t+c$
$=\tan (7-4 x)+c$
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