When a change of variable results in equal limits of integration

You haven't said what $f$ is. Once you fill in all the parts of the first integral $\int_a^b f(g(t))g'(t)dt$, you can check that the change of variable formula is true - for example, $$\int_0^\pi \sin(t)\cos(t)dt=0=\int_0^0 xdx$$

Usually, when you find that some substitution you've done has made the lower and upper limits identical, it's always a good idea to check if your integrand is symmetric or antisymmetric about the midpoint of the integration interval. For the specific case of

$$\int_0^\pi f(\sin(t))\cos(t)\,\mathrm dt$$

instead of doing the substitution $u=\sin\,t$ which sets both limits of integration identical, you might instead try the simpler substitution $t=v+\frac{\pi}{2}$, which turns your integral into

$$\int_{0-\frac{\pi}{2}}^{\pi-\frac{\pi}{2}} f\left(\sin\left(v+\frac{\pi}{2}\right)\right)\cos\left(v+\frac{\pi}{2}\right)\,\mathrm dv$$

or

$$-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f\left(\cos\,v\right)\sin\,v\,\mathrm dv$$

Now, noting that $f\left(\cos\,v\right)\sin\,v$ is odd ($f\left(\cos(-v)\right)\sin(-v)=-f\left(\cos v\right)\sin v$), the integral can be split like so

$$-\left(\int_{-\frac{\pi}{2}}^0 f\left(\cos\,v\right)\sin\,v\,\mathrm dv+\int_0^{\frac{\pi}{2}} f\left(\cos\,v\right)\sin\,v\,\mathrm dv\right)$$

and then we can do the following:

$$-\left(\int_{\frac{\pi}{2}}^0 f\left(\cos(-v)\right)(-\sin(-v))\,\mathrm dv+\int_0^{\frac{\pi}{2}} f\left(\cos\,v\right)\sin\,v\,\mathrm dv\right)$$

$$-\left(-\int_0^{\frac{\pi}{2}} f\left(\cos\,v\right)\sin\,v\,\mathrm dv+\int_0^{\frac{\pi}{2}} f\left(\cos\,v\right)\sin\,v\,\mathrm dv\right)$$

and you can now see that the integral is supposed to be zero, which is consistent with the result from the substitution that made both integration limits the same.