Question

Integrate the function ${\sec }^2(7-4x)$

Answer 1

Let $7-4x=t$
$\therefore -4dx=dt$
$\therefore \int {\sec }^2(7-4x)dx=\frac{-1}{4}\int se{c}^{2}tdt$
$=\frac{-1}{4}\left(\tan t\right)+C$
$=\frac{-1}{4}\tan (7-4x)+C$