Evaluate the following integrals:


Evaluate: $\int \sec ^{2}(7-4 x) d x$


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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