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Find the integrals of the function: sin2 (2x + 5) - Mathematics

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Question

Find the integrals of the function:

sin2 (2x + 5)

Sum
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Solution

Let `I = int sin^2 (2x + 5)` dx

  ....`[because sin^2 A = (1 - cos 2A)/2]`

`= 1/2 int [(1 - cos (4x + 10)]` dx

`= 1/2 int dx - 1/2 int cos (4x + 10)` dx

`= 1/2 [x - (sin (4x + 10))/4] + C`

Put 4x + 10 = t 

4 dx = dt

`x/2 - 1/2 . 1/4 int cos t  dt`

`= x/2 - 1/8  sin t + C`

`= x/2 - 1/8  sin (4x + 10) + C`

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Chapter 7: Integrals - Exercise 7.3 [Page 307]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 7 Integrals
Exercise 7.3 | Q 1 | Page 307

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