Question:
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?
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.
From your calculations, it appears there might be a mistake in the simplification process. Let's go through the steps again to verify the result.
Starting with the left-hand side (LHS) of the equation:
LHS = ∫[(x*sin2x*sin(π/2*cos x))/(2x-π)] dx from (0-π)
Using the trigonometric identity sin(2x) = 2*sin(x)*cos(x), we have:
LHS = ∫[(x*2*sin(x)*cos(x)*sin(π/2*cos x))/(2x-π)] dx from (0-π)
Expanding the term sin(π/2*cos x) using the same identity sin(2x) = 2*sin(x)*cos(x), we get:
LHS = ∫[(x*2*sin(x)*cos(x)*[2*sin(π/2*cos x)*cos(π/2*cos x)])/(2x-π)] dx from (0-π)
Simplifying further:
LHS = ∫[(4x*sin(x)*cos(x)*sin(π/2*cos x)*cos(π/2*cos x))/(2x-π)] dx from (0-π)
At this point, we can simplify the numerator by using the identity sin(a)*cos(b) = (1/2) * [sin(a+b) + sin(a-b)]. Applying this to the above expression:
LHS = ∫[(4x*sin^2(x)*[sin((π/2)*cos x + π/2*cos x) + sin((π/2)*cos x - π/2*cos x)])/(2x-π)] dx from (0-π)
LHS = ∫[(4x*sin^2(x)*[sin((π/2)*cos x + π/2*cos x) + sin((π/2)*cos x - π/2*cos x)])/(2x-π)] dx from (0-π)
Simplifying the sine terms:
LHS = ∫[(4x*sin^2(x)*[sin(π*cos x) + sin(0)])/(2x-π)] dx from (0-π)
LHS = ∫[(4x*sin^2(x)*sin(π*cos x))/(2x-π)] dx from (0-π)
Now, to prove that this is equal to the right-hand side (RHS) of the equation:
RHS = ∫[sin(x)*cos(x)*sin(π/2*cos x)] dx from (0-π)
Notice that the numerator in the LHS expression is equal to twice the numerator in the RHS expression. Therefore, we can directly see that:
LHS = 2 * RHS
Thus, the original equation you provided is incorrect, and there is no need to proceed further with the simplification.