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Question

Evaluate sin2(2x+5)dx.

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Solution

I=sin2(2x+5)dx

I=1cos2(2x5)2dx .......... ( 1cos2A=2sin2A)

I=121cos(4x+10)dx

I=12[xsin(4x+10)4+C]

I=x218sin(4x+10)+C

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