# 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.

• Yes. The solution would not even need the concept of $\pi$. Jan 10, 2019 at 16:52
• If you evaluate the integrals using contour integration methods (which doesn't involve computing the primitive function as we do when using ordinary calculus methods), then the factor of 1/8 can be seen to come from the coefficient of 1/x in the large x expansion of the square root, so it's a binomial coefficient. Jan 10, 2019 at 17:01
• I think you should include a little more emphasis on the "algebraic means" part of this question. As you might've seen, I posted a solution (which I since deleted) which used the Gamma function, as I was unclear what you were requesting. Jan 10, 2019 at 18:18
• @CountIblis Contour integration is likely well beyond the scope here. See my posted solution for a way forward using elementary calculus only. ;-)) Jan 10, 2019 at 18:18

## 2 Answers

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!

• I admit that your solution is much better for this problem (+1) than mine, but I disagree that the use of the Beta function is overkill in this situation. It is incredibly efficient and can be applied to both problems, effectively killing two birds with one stone. Why not? Jan 10, 2019 at 18:21
• @clathratus Use of the Beta function is fine. But then you need to show that $B(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$ and you need to show that $\Gamma(1/2)\sqrt \pi$. Those require more work that one needs and some of the steps involve carrying out integrals, which we could easily do here in a straightforward way. Finally, the OP has edited to state that use of the Gamma function is prohibited. Jan 10, 2019 at 18:35

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 $$A=8B.$$