How to proof this equation given a substitution?

Question

I need to proof that $$\int_{0}^{\pi}xf(\sin(x))dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin(x))dx$$ The problem says that the substitution $u=\pi-x$ is useful. Following their advice $u=\pi-x,du=-dx$ $$-\int_{0}^{\pi}(\pi-u)f(\sin(\pi-u))du$$ What do I do next, this the first time I encounter integrals writeen like this and I am not that familiar with proof thanks in advance !

Answer 1

\begin{align*} I&=\int_{0}^{\pi}(\pi-u)f(\sin(\pi-u))du\\ &=\int_{0}^{\pi}(\pi-u)f(\sin u)du\\ &=\pi\int_{0}^{\pi}f(\sin u)du-\int_{0}^{\pi}uf(\sin u)du\\ &=\pi\int_{0}^{\pi}f(\sin u)du-I. \end{align*}

Answer 2

$$\int_{0}^{\pi}xf(\sin(x))dx=$$ $$\int_{0}^{\pi}(\pi-x)f(\sin(\pi-x))dx=$$

$$\int_{0}^{\pi}(\pi-x)f(\sin x)dx=$$ $$\pi\int_{0}^{\pi}f(\sin x)du-\int_{0}^{\pi}xf(\sin x)dx \implies $$ $$\int_{0}^{\pi}xf(\sin(x))dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin(x))dx$$