Simple proof needed

[John O' Meara]

Prove that \int_0^a f(x)dx = \int_0^{\frac{a}{2}}[f(x) + f(a-x)]dx \\
Hence prove that \int_0^{\pi} x sin^6x dx = \pi \int_0^{\frac{\pi}{2}} sin^6x dx \\ and evaluate this integral using the following reduction formula I_{m,n}= \frac{n-1}{m+n}I_{m,n-2} \\

My effort:
\int_0^a f(x)dx = \int_0^{\frac{a}{2}}f(x)dx + \int_{\frac{a}{2}} f(a-x)dx \\ . Now
\int_{\frac{a}{2}}^a f(x)dx = \int_{\frac{a}{2}}^a f(a-x)dx \\ ... by a theorem. Therefore \int_0^a f(x)dx = \int_0^{\frac{a}{2}} f(x)dx + \int_{\frac{a}{2}}^a f(a-x)dx \\ . I need help to change the integral's limits to 0 and a/2 on the second integral on the right, thanks very much.

[MathematicalPhysicist]

\int_{0}^{a}f(x)dx
change variable u=a-x then we have du=-dx
-\int_{a}^{0}f(u)du=-\int_{0}^{a}f(a-x)dx
so now you have: \int_{0}^{a}f(x)dx=\int_{0}^{a}[f(x)-f(a-x)]/2dx
another change u=x-(a/2), which gives you:

\int_{-a/2}^{a/2}f(u+a/2)-f(a/2-u)]/2du

this function in the integrand is an odd function, which means that this equals 2 times the integral but from 0 to a/2.
if im mistaken, sorry for misleading.

[Gib Z]

You are mistaken :( \int^a_0 f(x) dx = \int^a_0 f(a-x) dx

On your 2nd line of latex, Reversing the bounds as you did makes the integrals value negative, canceling the other negative.

Jean - To finish it off, u=x - (a/2) for the last integral.

[MathematicalPhysicist]

well it should be -\int_{a}^{0}f(a-u)du=\int_{0}^{a}f(a-u)du
so yes, you are right, thanks for the correction.