How to prove this integral equation?

[Ganesh Ujwal]

How to prove that $\int_0^\pi {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi {f(\sin x)} \,dx$

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]?

 

[alyafey22]

How to prove that $\int_0^\pi {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi {f(\sin x)} \,dx$

It might be easier to change the problem to proving

$$\int_0^\pi \left(\frac{\pi}{2}-x \right)\,f(\sin x)\, dx = 0 $$

 

[mathmari]

$$\int_0^\pi {x\,f(\sin x)} \,dx =\frac{\pi}{2}\int_0^\pi {f(\sin x)} \,dx $$$$\int_0^\pi {x\,f(\sin x)} \,dx =\int_0^\pi {x\,f(\sin (\pi-x))} \,dx $$

$u=\pi-x \Rightarrow dx=-dx$

$x=0 \Rightarrow u=\pi, x=\pi \Rightarrow u=0$

$$\int_0^\pi {x\,f(\sin x)} \,dx =\int_0^\pi {x\,f(\sin (\pi-x))} \,dx =-\int_\pi^0 {(\pi-u)\,f(\sin u)} \,du \\ =\int_0^\pi {(\pi-u)\,f(\sin u)} \,du=\pi \int_0^\pi {\,f(\sin u)} \,du-\int_0^\pi {u\,f(\sin u)} \,du=\pi \int_0^\pi {\,f(\sin x)} \,dx-\int_0^\pi {x\,f(\sin x)} \,dx \\ \Rightarrow \int_0^\pi {x\,f(\sin x)} \,dx=\pi \int_0^\pi {\,f(\sin x)} \,dx-\int_0^\pi {x\,f(\sin x)} \,dx \\ \Rightarrow 2 \int_0^\pi {x\,f(\sin x)} \,dx=\pi \int_0^\pi {\,f(\sin x)} \,dx \\ \Rightarrow \int_0^\pi {x\,f(\sin x)} \,dx=\frac{\pi}{2}\int_0^\pi {\,f(\sin x)} \,dx$$