I'm trying to understand how this solution solves this question part ii). The solution does a substitution of $y=\pi -x$ into $\int^\pi_0 x f(\sin x) dx$ This substitution results in $$ \int^\pi_0 x f( \sin x)dx = \pi \int^\pi_0 f(\sin y) dy - \pi \int^\pi_0 y f(\sin y)dy$$
Then the author simply substitutes y = x to get : $$ \int^\pi_0 x f( \sin x)dx = \pi \int^\pi_0 f(\sin x) dx - \pi \int^\pi_0 y f(\sin x)dx$$ Which becomes $$ 2\int^\pi_0 x f( \sin x)dx = \pi \int^\pi_0 f(\sin x) dx $$
Doesn't subbing $y = x$ contradict the initial substitution of $y = \pi - x$ ? I've looked through my u-substitution textbook and nothing like this comes up. If someone could explain to me what this concept is called plus other examples of it, that would be great. Thank you.
A note in the solution you have provided refers to something called a dummy variable. For example, any integral that can be written in the form $$\int_a^b f(\lambda)\,d\lambda$$ will have the same value (and $\lambda$ can be any variable).
Therefore, in your solution, we have \begin{align} \int_0^\pi xf(\sin x)\,dx&=\int_\pi^0(\pi-y)f(\sin(\pi-y))(-1)\,dy\\ &=\int_0^\pi (\pi-y)f(\sin y)\,dy\\ &=\pi\int_0^\pi f(\sin \color{blue}{y})\,d\color{blue}{y}-\int_0^\pi \color{blue}{y}f(\sin \color{blue}{y})\,d\color{blue}{y}\\ \int_0^\pi xf(\sin x)\,dx&=\pi\int_0^\pi f(\sin \color{blue}{x})\,d\color{blue}{x}-\int_0^\pi \color{blue}{x}f(\sin \color{blue}{x})\,d\color{blue}{x}\\ 2\int_0^\pi xf(\sin x)\,dx&=\pi\int_0^\pi f(\sin x)\,dx\\ \int_0^\pi xf(\sin x)\,dx&=\frac\pi2\int_0^\pi f(\sin x)\,dx\\ \end{align} as desired.