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Question
Find the integrals of the function:
sin3 (2x + 1)
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Solution
Let `I = int sin^3 (2x + 1)` dx
`= 1/4 [3 sin (2x + 1) - sin 3 (2x + 1)]` dx
.....`[because sin^3 theta = 1/4 (3 sin theta - sin 3 theta)]`
`= 3/4 (- (cos (2x + 1))/2 - 1/4 ((- cos 3 (2x + 1))/6) + C`
`= -3/8 cos (2x + 1) + 1/24 cos 3 (2x + 1) + C`
`= - 3/8 cos (2x + 1) + 1/24 [4 cos^3 (2x + 1) - 3 cos (2x + 1) + C` ......`[because cos 3 theta = 4 cos^3 theta - 3 cos theta]`
`= - 3/8 cos (2x + 1) + 1/6 cos^3 (2x + 1) - 1/8 cos(2x + 1) + C`
`= - 1/2 cos (2x + 1) + 1/6 cos^3 (2x + 1) + C`
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