MathematicsIntegration
posted by Shenaya .
Question:
Prove that [integrate {x*sin2x*sin[π/2*cos x]} dx] /(2xπ) } from (0π)
= [ integrate {sin x*cos x*sin[π/2*cos x} dx ] from (0π).
My thoughts on the question:
We know that integrate f(x) dx from (0a) = integrate f(ax) dx from (0a)
From that and by sin(2x)=2sin(x)*cos(x)
L.H.S. = integrate { (πx)*2sin(πx)*cos(πx)*sin[(π/2)cos(πx)] dx] /[2(πx)  x]}from (0π)
= integrate { [ (πx)*2sinx*cosx *[ sin(π/2*coss x] dx ]/(π2x)} from (0π)
= integrate { [(πx) 2sinx*cosx*[sin(π/2*cosx] dx/(π2x) } from (0π)
But in the result they are asking is to prove,there's no "2" and terms of (πx) and (π2x)
Did I made a mistake?

pi is displayed as a question mark here.

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). 
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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