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Question

Evaluate:π20cos2xdx

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Solution

π20cos2xdx
=12π202cos2xdx
We know that cos2x=2cos2x12cos2x=1+cos2x
=12π20(1+cos2x)dx
=12π20dx+12π20cos2xdx
=12[x]π20+12[sin2x2]π20
=12[x]π20+14[sin2x]π20
=12[π20]14[sinπsin0]
=π40
=π4

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