Comparison of integrals by algebraic means $$ \begin{align}A&:=\int_0^1\frac1{\sqrt{x(1-x)}}\ \mathrm dx \\ B&:=\int_0^1\sqrt{x(1-x)}\ \mathrm dx \end{align} $$
My CAS tells me that $A = \pi$ and $B = \frac18\pi$.
How can one prove that $A=8B$ using just basic rules of integration such as the chain rule?
Trigonometric functions are not allowed since they are not definable as integrals. Neither is the Gamma function allowed, since it is defined in terms of exp, which is like a trigonometric function. These restrictions are part of what I mean by "algebraic means". On the other hand, integration by parts is fine. Equivalently, the fundamental theorem of calculus is also fine.


If integration by parts is an acceptable approach, then we can proceed as follows.


First, let $B$ be the integral defined as
$$B=\int_0^1 \sqrt{x(1-x)}\,dx\tag1$$
Integrating by parts with $u=\sqrt{x(1-x)}$ and $v=x$ in $(1)$, we obtain
$$B=\frac12 \int_0^1 x\left(\frac{\sqrt x}{\sqrt{1-x}}-\frac{\sqrt{1-x}}{\sqrt x}\right)\,dx\tag2$$ 

Now enforcing the substitution $x\mapsto 1-x$ in the first term on the right-hand side of $(2)$ reveals 
$$\int_0^1 x\frac{\sqrt x}{\sqrt{1-x}}\,dx=\int_0^1 \frac{\sqrt{1-x}}{\sqrt x}\,dx-\int_0^1 \sqrt{x(1-x)}\,dx\tag3$$

Substituting $(3)$ into $(2)$ we find that 
$$B=\frac14 \int_0^1 \frac{\sqrt {1-x}}{\sqrt{x}}\,dx\tag 4$$

Finally, integrating by parts with $u=\frac{\sqrt {1-x}}{\sqrt{x}}$ and $v=x$ in $(4)$ yields
$$B=\frac18 \int_0^1 \frac{1}{\sqrt{x(x-1)}}\,dx$$
as was to be shown!
