Question

Evaluate : ∫tan²(2x – 3)dx

Answer 1

Let I = ∫tan²(2x – 3)dx

= ∫tan²(2x – 3)dx

= ∫sec²(2x – 3) - 1 dx

Let 2x -3 = t dx = dt/2

= \(\frac{1}{2}\int\)sec²t - 1 dt

= \(\frac{1}{2}\)tan t –x

Substitute the value of t.

Hence,

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