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Question

Find the integrals of the functions.
sin3(2x+1)dx

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Solution

sin3(2x+1)dx=sin2(2x+1).sin(2x+1)dx=[1cos2(2x+1)]sin(2x+1)dx (sin2x=1cos2x)
Let cos(2x+1)=t2sin(2x+1)dx=dtsin(2x+1)dx=dt2
sin3(2x+1)dx=12(1t2)dt=12(tt33)+C=12(cos(2x+1)cos3(2x+1)3)+C=cos(2x+1)2+cos3(2x+1)6+C


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