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Integrate the functions: tan2(2x – 3)

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प्रश्न

Integrate the functions:

tan2(2x – 3)

योग
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उत्तर

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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अध्याय 7: Integrals - Exercise 7.2 [पृष्ठ ३०५]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.2 | Q 21 | पृष्ठ ३०५

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