TITLE: Comparison of integrals by algebraic means
QUESTION [13 upvotes]: $$ \begin{align}A&:=\int_0^1\frac1{\sqrt{x(1-x)}}\ \mathrm dx \\ B&:=\int_0^1\sqrt{x(1-x)}\ \mathrm dx \end{align} $$
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.

REPLY [7 votes]: In fact, by integration by parts, one has
\begin{eqnarray*}
B&=&\int_0^1\sqrt{x(1-x)} \mathrm dx\\
&=&-\frac12\int_0^1\frac{x(1-2x)}{\sqrt{x(1-x)}}\mathrm dx\\
&=&-\frac12\int_0^1\frac{x(1-x)-x^2}{\sqrt{x(1-x)}}\mathrm dx\\
&=&-\frac12B+\frac12\int_0^1\frac{x^2}{\sqrt{x(1-x)}}\mathrm dx
\end{eqnarray*}
and hence
$$ 3B=\int_0^1\frac{x^2}{\sqrt{x(1-x)}}\mathrm dx. \tag{1}$$
Under $1-x\to x$, (1) becomes
$$ 3B=\int_0^1\frac{(1-x)^2}{\sqrt{x(1-x)}}\mathrm dx. \tag{2}$$
Adding (1) to (2), one has
$$ 6B=\int_0^1\frac{x^2+(1-x)^2}{\sqrt{x(1-x)}}\mathrm dx=\int_0^1\frac{1-2x(1-x)}{\sqrt{x(1-x)}}\mathrm dx=A-2B.$$
This implies
$$ A=8B. $$