Comparison of integrals by algebraic means

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

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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

$$B=\frac12 \int_0^1 x\left(\frac{\sqrt x}{\sqrt{1-x}}-\frac{\sqrt{1-x}}{\sqrt x}\right)\,dx\tag2$$

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Now enforcing the substitution $x\mapsto 1-x$ in the first term on the right-hand side of $(2)$ reveals

$$\int_0^1 x\frac{\sqrt x}{\sqrt{1-x}}\,dx=\int_0^1 \frac{\sqrt{1-x}}{\sqrt x}\,dx-\int_0^1 \sqrt{x(1-x)}\,dx\tag3$$

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Substituting $(3)$ into $(2)$ we find that

$$B=\frac14 \int_0^1 \frac{\sqrt {1-x}}{\sqrt{x}}\,dx\tag 4$$

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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!

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 $$ A=8B. $$