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Question
Integrate the functions:
tan2(2x – 3)
Solution
Let `I = int tan^2 (2x - 3) dx`
`= int [sec^2 (2x - 3) - 1]dx`
`= int sec^2 (2x - 3)dx - int 1 dx`
`= sec^2 (2x - 3) dx - x + C_1`
I = I1 - x + C1
Where, `I_1 = int sec^2 (2x - 3) dx.`
Put 2x - 3 = t
⇒ 2dx = dt
⇒ `I_1 = 1/2 int sec^2 t dt`
⇒ `I_1 = 1/2 tan t + C_2`
`= 1/2 tan (2x - 3) + C_2`
`I = I_1 - x + C_1`
= `1/2 tan (2x - 3) - x + C`
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