Question

Intergrate by substitution method

Answer 1

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

on integrating

tan(2x-3)/2 + x + c

hope it helped you

Answer 2

u=2x-3

du=2 dx

dx=1/2 du

∫tan²(2x-3) dx

$\frac{1}{2}$∫tan² u du

$\frac{1}{2}$( ∫sec²u du -∫1 du)

$\frac{1}{2}$ (tan u - u )

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

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

3/2 will add to constant

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

$\frac{1}{2}$ tan(2x-3) -x + λ