When a change of variable results in equal limits of integration Here is something that's been bugging me for a while. Say we want to find $$I = \displaystyle\int\nolimits_{a}^{b}f(g(t))g'(t)\,\mathrm dt.$$ 
If we substitute $x = g(t)$, then $$I = \displaystyle\int_{g(a)}^{g(b)}f(x)\,\mathrm dx.$$ But what do we do in the case where $g(a) = g(b)$? How do we interpret that? 
For example, if $x = \sin{t}$, $a = 0$, and $b = \pi$, then $g(a) = \sin(0) = \sin(\pi) = g(b)$. It would appear that the integral would be zero, but I find it isn't. 
What to do when a given substitution makes both limits the same?
 A: 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, $$\int_0^\pi \sin(t)\cos(t)dt=0=\int_0^0 xdx$$
A: 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.
