Given the substitution, prove the integral is true

Question

Given that u = Pi - x , prove that the integral from 0 to Pi of x*f(Sin(x)) dx is equal to (Pi/2) * the integral from 0 to Pi of f(Sin(x)) dx

 

I took the derivative of U and got du = 0-1dx, which I changed to -1*du = dx. Then I changed u = Pi - x to Pi - U = x, then substituted that in to the right equation. From there I changed the start and end integral points from 0 and Pi, to Pi and 0 by using the substitution formula. By making the integral negative, I can put them in the right order again, and when I factor out the -1 that occurs in the next section, I end up with the integral from 0 to Pi of (Pi-U)*f(Sin(Pi-u))du. 

 

From here I used the identity sin(a-b) = sin(a)cos(b)-sin(b)cos(a) on the interior sin(Pi-U) and then use the negative to fix the integral like I stated above.

 

Now I have the integral from 0 to Pi of (Pi - u)*f(sin(u)) du and don't know how to proceed.

Answer 1

An alternate method is to consider

 

  the integral  0 to pi of (x - pi/2) f(Sin(x))       This integral can be shown to be zero (proof outlined below)

   Thus the integral  0 to pi of x f(Sin(x)) =  pi/2  times integral 0 to pi of f(Sin(x))   QED

 

  why is  integral 0 to pi of (x - pi/2) f(Sin(x))   = 0   ?

 

  Basically because  about the midpoint  of the interval [0, pi],   (x -pi/2) is in odd function while

    f(Sin(x)) is an even function.   The integral of the product of an odd function and and even function over an interval symmetrically placed about the point about which even and odd are predicated is zero.

   This can be seen more explicitly by shifting the integration from 0 to pi to  -pi/2 to + pi/2