Question

integrate the function tan²(2x - 3).dx

Answer 1

given, I = ∫tan²(2x - 3)dx

= ∫[sec²(2x - 3) - 1]dx

= ∫sec²(2x - 3)dx - ∫dx

= ∫sec²(2x - 3)dx - x + C₂

let I₁ = ∫sec²(2x - 3)dx

let (2x - 3) = t

differentiate both sides,

2dx = dt ⇒ dx = 1/2 dt

I₁ = 1/2 ∫sec²t.dt

= 1/2 tant + C₁

put (2x - 3) = t

I₁ = tan(2x - 3) + C₁

Hence, I = ∫sec²(2x - 3)dx - x + C₂

= 1/2tan(2x - 3) + C₁ - x + C₂

= 1/2tan(2x - 3) - x + (C₁ + C₂)

= 1/2tan(2x - 3) - x + C [where, C = C₁ + C₂ ]

Answer 2

Hope it will help you!!