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Question

Evaluate: sin2xdx

A
x2sin2x4+C
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B
x2sin2x2+C
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C
x4sin2x4+C
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D
x2sin2x3+C
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Solution

The correct option is A x2sin2x4+C
sin2xdx=1cos2x2dx (sin2x=1cos2x2)
=12(1cos2x)dx
=12dx12cos2xdx
=x2+(12)sin2x2+C=x2sin2x4+C

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