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