Samuel H.

# Given the substitution, prove the integral is true

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.

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