Question

Integrate the function: $tan^ 2 (2x -3)$

Answer 1

$tan^2 (2x -3) = sec^2 (2x -3) -1$

Let 2x - 3 = t

∴ 2dx = dt

$\int \tan^2(2x-3)dx=\int\bigg[(\sec^2(2x-3))-1\bigg]dx$

=$\frac{1}{2}\int(\sec^2 t)dt-\int 1dx$

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

= $\frac{1}{2}\tan t -x+C$

= $\frac{1}{2}\tan(2x-3)-x+C$