It is obvious without taking an antiderivative that $$\tag 1\int_0^\pi \cos\theta\,d\theta =0.$$ But suppose we do the tangent half-angle substitution that is used for rational functions of sine and cosine (often incorrectly called the Weierstrass substitution, partly but not entirely because James Stewart is better at mathematics than at history) we get this: $$ \int_0^\infty \frac{1-x^2}{1+x^2} \cdot \frac{2\,dx}{1+x^2}. \tag 2 $$ Now suppose we were confronted with $(2)$ and we're not clever enough to realize it's reducible to $(1),$ and we use the substitution $$ w = \frac 1 x. $$ We get $$ \int_0^\infty \frac{1-x^2}{1+x^2} \cdot \frac{2\,dx}{1+x^2} = \int_\infty^0 \frac{1-w^2}{1+w^2} \cdot \frac{2\,dw}{1+w^2}, \tag 3 $$ and solving $I=-I$ we get $I=0.$
My question is whether there is some way of looking at $(2)$ that makes it obvious that it is $0$ in the way in which it is obvious that $(1)$ is $0,$ without noticing that $(2)$ can be transformed to $(1)$ or otherwise using transcendental functions? Maybe some geometric interpretation? Line $(3)$ doesn't seem obviously true; you have to do some work even though that work is trivial.
On $[0,\infty)$, the denominator of (2) is positive. The numerator of (2) is positive on $[0,1)$ and negative on $(1,\infty)$. Maybe we can get $\int_0^1 = -\int_1^\infty$, possibly cancelling "pointwise", although "pointwise" needs some way to associate points of $(1,\infty)$ to points of $(0,1)$. The simplest way to map $(1,\infty)$ to $(0,1)$ is $z \mapsto 1/w$. If we try that, we get exactly what we hoped to get -- the negative and positive parts are the same integral (but for a minus sign).
(This answer seems like a cheat to me. I won't be surprised if others think so as well.)