STEP 2 Calculus module Q2

2006 S2 Q4
For the intro it says show that
integral(xf(sinx)dx = pi/2*integral(f(sinx)dx
upper limit = pi, lower limit = 0 on both integrals.
hint says you can do this by using a substitution x=pi - t and then proceeds by swapping the t for x.

But then we would get pi = 0 which is nonsense?
I don't get these substitutions?

2006 S2 Q4
For the intro it says show that
integral(xf(sinx)dx = pi/2*integral(f(sinx)dx
upper limit = pi, lower limit = 0 on both integrals.
hint says you can do this by using a substitution x=pi - t and then proceeds by swapping the t for x.

But then we would get pi = 0 which is nonsense?
I don't get these substitutions?

$\begin{aligned} \int_0^{\pi} x f(\sin x) \ dx & = -\int_{\pi}^{0} (\pi - t) f(\sin [\pi - t]) \ dt \\[0.3cm] & = \pi \int_0^{\pi} f(\sin t) \ dt - \int_0^\pi tf(\sin t) \ dt \end{aligned}$

Since $t$ is a dummy variable we can replace it with $x$. Another way to think about it is that the two integrals on the RHS are in terms of $t$ but if we replace it with $x$ then the values of the integrals will not change! Since they are definite integrals, we say that the $t$ 'integrates out' because it goes away at the end of process (i.e. you replace it with limits!) which means the $t$ doesn't stick around and we can replace it with any other variable while maintaining the value of the integral.

Anyway, we get



from which point onward the final result is obtained via rearrangement.

So if they are indefinite integrals can I still do the substitution exchange