Question

Solve: $\int {\tan }^2(2x-3)dx$

Answer 1

Formula:

$\int se{c}^{2}xdx=\tan x+c$

${\tan }^{2}x={\sec }^{2}x-1$

$\int {\tan }^2(2x-3)dx=\int \left[{\sec }^2\right(2x-3)-1]dx$

$=\int {\sec }^2(2x-3)dx-\int 1dx$

$=\tan (2x-3)\frac{d}{dx}(2x-3)-x+c$

$=\tan (2x-3)\left(2\right)-x+c$

$=2\tan (2x-3)-x+c$ (Ans)