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Question

Evaluate : π/20cos3xdx

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Solution

I=π20cos3xdx

I=π20cos2x.cosxdx=π20(1sin2x).cosxdx

Let u=sinx;dx=cosxdx

I=π20(1u2)dx=[uu33]π20

I=[sinxsin3x3]π20=113

I=23.

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