Question

Integrate the function: $sec^2 (7 -4x)$

Answer 1

Let 7 - 4x = t

∴ -4dx = dt

∴ $\int \sec^2(7-4x)dx=\frac{-1}{4}\int sec^2tdt$

=$\frac{-1}{4}(\tan\, t)+C$

= $\frac{-1}{4}\tan(7-4x)+C$