Question:
Integrate tan^2 (2x-3)
Answer:
Given ∫tan2 (2x-3)dx
= ∫{sec2 (2x-3) - 1}dx (since tan2 x = sec2 x - 1 )
=> [{tan(2x - 3)}/2 - x] + c (∫sec2 x dx = ∫tan2 x )
So ∫tan2 (2x-3)dx = [{tan(2x - 3)}/2 - x] + c
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