Prove that [integrate {x*sin2x*sin[π/2*cos x]} dx] /(2x-π) } from (0-π)
= [ integrate {sin x*cos x*sin[π/2*cos x} dx ] from (0-π).

My thoughts on the question:

We know that integrate f(x) dx from (0-a) = integrate f(a-x) dx from (0-a)

From that and by sin(2x)=2sin(x)*cos(x)

L.H.S. = integrate { (π-x)*2sin(π-x)*cos(π-x)*sin[(π/2)cos(π-x)] dx] /[2(π-x) - x]}from (0-π)
= integrate { [ (π-x)*2sinx*cosx *[ sin(π/2*coss x] dx ]/(π-2x)} from (0-π)
= integrate { [(π-x) 2sinx*cosx*[sin(π/2*cosx] dx/(π-2x) } from (0-π)

But in the result they are asking is to prove,there's no "2" and terms of (π-x) and (π-2x)

Did I made a mistake?

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asked by Shenaya
  1. pi is displayed as a question mark here.

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    posted by Shenaya
  2. I see π being properly displayed probably due to encoding.

    I also see that there is a removable discontinuity at x=π/2.

    Numerical integration (skipping x=π/2) gives identical results for both expressions, so hopefully no typo.
    Try translating limits to (-π/2, π/2).

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    posted by MathMate
  3. I've heard about "numerical integration" but we haven't been taught that.So I should solve this without using that method.

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    posted by Shenaya

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