I'm stuck on the second Step II calculus question.

Making the substitution of $x=\pi-t$, the integral becomes \[\int^0_\pi(\pi-t)f(\sin(\pi-t))\,(-dt)\] which tidies up to be \[\int^\pi_0(\pi-t)f(\sin t)\,dt\] I then split the integral into two parts: \[\pi\int^\pi_0f(\sin t)\,dt-\int^\pi_0tf(\sin t)\,dt\] and then subbing $x$ back in: \[\pi\int^\pi_0f(\sin x)\,dx-\int^\pi_0(\pi-x)f(\sin x)\,dx\] which obviously just cancels out the original substitution.

The hints sheet seems to imply that I can just change all the $t$'s to $x$'s (which would then give the required result), but I don't think that could be right because $x\neq\pi-x$.