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Question

Evaluate :
π/20cos2x dx

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Solution

Consider I=π/20cos2x dx .........(1)

I=π/20cos2(π2x)dx (a0f(x)dx=a0f(ax)dx)

I=π/20sin2x dx .....(2)

Adding 1 and 2, we get,

2I=π/20dx

2I=π20

I=π4

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