Question

Integrate the function ${\tan }^2(2x-3)$

Answer 1

${\tan }^2(2x-3)={\sec }^2(2x-3)-1$
Let $2x-3=t$
$\therefore 2dx=dt$
$\Rightarrow \int {\tan }^2(2x-3)dx=\int \left[\right({\sec }^2(2x-3))-1]dx$
$=\frac{1}{2}\int \left({\sec }^{2}t\right)dt-\int 1dx$
$=\frac{1}{2}\int {\sec }^{2}tdt-\int 1dx$
$=\frac{1}{2}\tan t-x+C$
$=\frac{1}{2}\tan (2x-3)-x+C$