Question

Evaluate the given integral.
$\int {\sec }^2(7-4x)dx$

Answer 1

Let $t=7-4x\Rightarrow dt=-4dx$

$\Rightarrow \frac{-dt}{4}=dx$

$I=\int {\sec }^2(7-4x)dx$

$=\frac{-1}{4}\int {\sec }^{2}tdt$

Let $u=\tan t\Rightarrow dt={\sec }^{2}dt$

$=\frac{-1}{4}\int du$

$=\frac{-1}{4}u+c$ ..........where $c$ is the constant of integration.

$=\frac{-1}{4}\tan t+c$ ............where $u=\tan t$

$=\frac{-1}{4}\tan (7-4x)+c$ ............where $t=7-4x$