One might solve the integral $\int_0^1 \sqrt{1-x^2}\,\mathrm{d}x$ by invoking the substitution $x \mapsto \sin x$. This gives the standard integral $\int_0^{\pi/2} \cos^2(\theta) \,\mathrm{d}\theta$. A substition can be viewed as a mapping between two coordinate systems, where the only thing that is constant is the area. Assume that we want to transform $\sqrt{1-x^2}$ on $x\in[0,1]$ continuously into $\cos^2x$ on $x\in[0,\pi/2]$. One way to do this is to look at $$ g(x,T) = \bigl((1-T)\cdot 1 + T \cdot \cos x \bigr) \sqrt{ 1 -\bigl( (1-T)x + T \sin x \bigr)^2} $$ Where $T$ ranges from $0$ to $1$. This is in my eyes a linear transform between the two functions. Meaning $g(x,0) = \sqrt{1 - x^2}$ and $g(x,1)=\cos^2x$, assuming we still only consider the area in the first quadrant) Assume that we integrate $g(x,T)$ from $0$ to $x^*$ where $x^*$ is defined such that $$1 = (1-T)x^* + T \sin x^*$$
My question is if this transform conserves the area. In other words is $$ \int_0^{x^*} g(x,T)\,\mathrm{d}x = \int_0^1 \sqrt{1-x^2}\,\mathrm{d}x = \int_0^{\pi/2} \cos^2x\,\mathrm{d}x $$ for all $T \in [0,1]$?
My numerical estimations shows that it is atleast extremely close. Assume that the assertion above fail to hold. Does there exists a continuous transformation from $\sqrt{1-x^2}$ to $\cos^2x$ in such a way that the area is preserved?
$$ g(x,T)\,\mathrm{d}x=\sqrt{1-u^2}\,\mathrm{d}u $$ where $u=(1-T)x+T\sin(x)$, so since $u=1$ when $x=x^\ast$, we have $$ \int_0^{x^\ast}g(x,T)\,\mathrm{d}x=\int_0^1\sqrt{1-u^2}\,\mathrm{d}u $$