How about calculating the area, $S$, of one quarter of the unit circle?
First note that $S = \int_0^1 {\sqrt {1 - x^2 } \,{\rm d}x} $. The substitution $x=\sin(u)$ (hence, ${\rm d}x = \cos(u) \,{\rm d}u$) gives $ \int_0^1 {\sqrt {1 - x^2 } \,{\rm d}x} = \int_0^{\pi /2} {\sqrt {1 - \sin ^2 (u)} \cos (u)\,{\rm d}u} = \int_0^{\pi /2} {\cos ^2 (u)\,{\rm d}u}. $ The integral on the right can be computed as follows. $ \int_0^{\pi /2} {\cos ^2 (u)\,{\rm d}u} = \int_0^{\pi /2} {\frac{{1 + \cos (2u)}}{2}\,{\rm d}u} = \frac{\pi }{4} + \frac{1}{2}\int_0^{\pi /2} {\cos (2u)\,{\rm d}u} = \frac{\pi }{4}, $ where the last equality follows from $ \int_0^{\pi /2} {\cos (2u)\,{\rm d}u} = \frac{1}{2}\int_0^\pi {\cos (t)\,{\rm d}t} = 0. $ Thus we have shown that $S=\pi/4$, which is a non-trivial result.
EDIT: As another example, using the the substitution $y=e^{-x}$, we have $ \int_0^\infty {e^{ - x} \,{\rm d}x} = \int_1^0 {y\frac{{\,{\rm d}y}}{{ - y}}} = \int_0^1 {1\,{\rm d}y} = 1. $