TITLE: How prove that $\int_0^\pi {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi {f(\sin x)} \,dx$
QUESTION [5 upvotes]: To prove that $\int_0^\pi  {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi  {f(\sin x)} \,dx$ is true, first I started calculating the integral of the left indefinitely
$$ \int {x\,f(\sin x)\,\,dx}  $$
using substitution:
$$ \sin x = t,\quad  x = \arcsin t, \quad {dx = \frac{{dt}}{{\sqrt {1 - {t^2}} }}}$$
is obtained:
$$ \int {x\,f(\sin x)\,\,dx} = \int {\arcsin t \cdot f(t) \cdot \frac{{dt}}{{\sqrt {1 - {t^2}} }}}$$
$$ \qquad\quad = \int {\frac{{\arcsin t\,dt}}{{\sqrt {1 - {t^2}} }} \cdot f(t)} $$
Then using integration by parts:
$$ \begin{array}{*{20}{c}}
  {u = f(t)},&{dv = \frac{{\arcsin t\,dt}}{{\sqrt {1 - {t^2}} }}} \\ 
  {du = f'(t)\,dt},&{v = \frac{{{{(\arcsin t)}^2}}}{2}} 
\end{array} $$
then:
\begin{align*}
  \int {x\,f(\sin x)\,\,dx}  &= f(t) \cdot \frac{{{{(\arcsin t)}^2}}}{2} - \int {\frac{{{{(\arcsin t)}^2}}}{2}}  \cdot f'(t)\,dt \\ 
   &= f(\sin x) \cdot \frac{{{x^2}}}{2} - \int {\frac{{{x^2}}}{2} \cdot f'(\sin x)\,\cos x\,dx}  \\ 
\end{align*}
Now, evaluating from 0 to $\pi$
\begin{align}
  \int_0^\pi  {x\,f(\sin x)} \,dx & = \left[ {f(t) \cdot \frac{{{x^2}}}{2}} \right]_0^\pi  - \int_0^\pi  {\frac{{{x^2}}}{2} \cdot f'(\sin x)\,\cos x\,dx} \\
  \int_0^\pi  {x\,f(\sin x)} \,dx & = f(0) \cdot \frac{{{\pi ^2}}}{2} - \int_0^\pi  {\frac{{{x^2}}}{2} \cdot f'(\sin x)\,\cos x\,dx} \qquad ..[1] \\ 
\end{align}
On the other hand, doing the same process with the integral on the right side I get:
\begin{equation}\int_0^\pi  {f(\sin x)} \,dx = f(0) \cdot \pi  - \int_0^\pi  {x \cdot f'(\sin x)\,\cos x\,dx}  \qquad ..[2] \end{equation}
And even here I do not have enough data to say that equality $\int_0^\pi  {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi  {f(\sin x)} \,dx$ is true. 
Can anyone suggest me what to do with the equalities [1] and [2]?
Thanks in advance.

REPLY [3 votes]: In general $$\int^b_a f(x) dx\,=\int^b_a f(b+a-x)dx\,$$