$u$-substitution tells us that $$ \int_{0}^{\pi/2}(\sin x)^2\cos x\,dx = \int_{0}^{1}u^2\,du = \frac{1}{3}, $$ where we made the substitution $u(x) = \sin x$, and $u(0)=0$, $u(\pi/2)=1$.
What I want to do is visualize a continuous transformation from one system of variables to the other, $$ u_t(x) = (1-t)\sin x +t x, \quad 0\leq t \leq 1, $$ where $u_0(x) = \sin x$, $u_1(x) = x$, and $$ \int_{a_t}^{b_t} (u_t(x))^2 u'_t(x) \,dx = \frac{1}{3}, \quad 0\leq t \leq 1. $$ The only trouble I'm having is figuring out what the bounds of the integrals should be, as a function of $t$. Obviously $a_0 = 0$, $b_0 =\pi/2$, and $a_1 = 0$, $b_1=1$, but I'm getting confused when $t$ is between $0$ and $1$.
Does what I'm looking for make sense? I've always thought of $u$-substitution (or change-of-variables) as changing one integral into the other by "stretching" or "deforming" the area under the graph. So naturally I though it should be possible to make that deformation explicit. Am I missing something obvious?
You want $$\int_{a_t}^{b_t} (u_t(x))^2 u'_t(x) \,dx = \frac {(u_t(x))^3}{3}\bigg|_{a_t}^{b_t} = \frac{1}{3}, \quad 0\leq t \leq 1.$$Thus you have to have $$(u_t(b_t))^3 -(u_t(a_t))^3=1$$