Question
$\int \sec ^2(7-4 x) d x=?$

Answer

$(b) \frac{-1}{4} \tan (7-4 x)+C$
Explanation: Given integral is $\int \sec ^2(7-4 x) d x=$ ?
Let, $7-4 x = z$
$\Rightarrow-4 dx=dz$
So,
$\int \sec ^2(7-4 x) dx=?$
$=\int \sec ^2 z \frac{d z}{-4}$
$=-\frac{1}{4} \int \sec ^2 z d z$
$\int \sec ^2(7-4 x) d x \text { where } c \text { is the integrating constant. }$
$=\int \sec ^2 z \frac{d z}{-4}$
$=-\frac{1}{4} \int \sec ^2 z d z$
$=-\frac{1}{4} \tan z+c$
$=-\frac{1}{4} \tan (7-4 x)+c$

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