By making the substitution $x=\pi -t$ show that; $$\int_{0}^{\pi} xf(\sin x)dx = \frac{1}{2}\pi \int_{0}^{\pi} f(\sin x)dx$$ where $f(\sin x) $ is a given function of $\sin x$.
I get $dx = -dt$ and that $\sin (\pi -t) = \sin t$ which gives me;
$$\int_{0}^{\pi} xf(\sin x)dx =\int_{0}^{\pi} tf(\sin t)dt - \pi \int_{0}^{\pi} f(\sin t)dt$$
Clearly I am missing a minus sign somewhere so where have I gone wrong?(note; the integral in t is equal to the integral in x, they're just variables)
You forgot to flip the limits. Letting $x=\pi -t$, we have $dx=(-1)\,dt$ and the limits are for $x=0$, $t=\pi$ and for $x=\pi$, $t=0$. Therefore, we have
$$\begin{align} \int_0^\pi xf(\sin x)\,dx&=\int_\pi^0 (\pi -t)f(\sin (\pi -t))\,(-1)\,dt\\\\&=\int_0^\pi (\pi - t)\,f(\sin t)\,dt\\\\ &=\pi \int_0^\pi f(\sin x)\,dx-\int_0^\pi x\,f(\sin x)\,dx\\\\ 2\int_0^\pi xf(\sin x)\,dx&=\pi \int_0^\pi f(\sin x)\,dx\\\\ \int_0^\pi xf(\sin x)\,dx&=\frac{\pi}{2}\int_0^\pi f(\sin x)\,dx \end{align}$$
As $t=\pi-x, x=0\iff t=\pi;x=\pi\iff t=0$
$$\int_{0}^{\pi} xf(\sin x)dx =\int_\pi^0(\pi-t)f(\sin(\pi-t))(-dt)$$
$$=-\int_\pi^0(\pi-t)f(\sin t)dt$$
$$=\int_0^\pi(\pi-t)f(\sin t)dt$$
$$=\pi\int_0^\pi f(\sin t)dt-\int_0^\pi tf(\sin t)dt$$
