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Question

If I=π/2π/2sin2x1+4x dx, then the value of 24πI is

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Solution

I=π/2π/2sin2x1+4x dx (1)
I=π/2π/2sin2(x)1+4x dx
I=π/2π/2sin2x1+4x dx
I=π/2π/24xsin2x1+4x dx (2)
Adding (1) and (2), we get
2I=π/2π/2sin2x dx
2I=2π/20sin2x dx
I=π/20sin2x dx=π4

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