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I was reading Advanced Integration Techniques, and found that$$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx =\frac{\pi}{8}$$

The book provides one method using residue theorem and Laurent expansion. However, I wonder if there are other techniques that I can use to evaluate this integral.

The most direct method would be solving the integral and plug in the limits. $$\int_{0}^{1}\sqrt{x}\sqrt{1-x}\,\mathrm dx = \left[\frac{\arcsin(2x-1)+\sqrt{(1-x)x}(4x-2)}{8}\right]_{0}^{1}=\frac{\pi}{8}$$

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    $\begingroup$ Beta function?${}$ $\endgroup$ – Lord Shark the Unknown Dec 31 '18 at 18:16
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    $\begingroup$ Maybe you know it, but it's worth to note that it's a particular case of the beta integral. $\endgroup$ – Madarb Dec 31 '18 at 18:17
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    $\begingroup$ $\sqrt{x}\sqrt{1-x} = \sqrt{x-x^2} = \sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}$; substitute $u = x-\frac{1}{2}$. $\endgroup$ – rogerl Dec 31 '18 at 18:18
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    $\begingroup$ $B\left( \frac 32,\frac 32\right)$ ? $\endgroup$ – Darkrai Jan 1 at 4:18
  • $\begingroup$ Well, solving a more general problem we can show that (when $\text{n}>0$): $$\mathcal{I}_\text{n}:=\int_0^\text{n}\sqrt{x}\cdot\sqrt{\text{n}-x}\space\text{d}x=\frac{\pi}{8}\cdot\text{n}^2\tag1$$ $\endgroup$ – Jan Jan 1 at 16:14
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Write $x=\sin^2 t$ so $dx=2\sin t\cos t dt$. Your integral becomes $$\int_0^{\pi/2}2\sin^2 t\cos^2 t dt=\int_0^{\pi/2}\frac12 \sin^2 2t dt=\int_0^{\pi/2}\frac{1-\cos 4t}{4}dt=\left[\frac{t}{4}-\frac{1}{16}\sin 4t\right]_0^{\pi/2}=\frac{\pi}{8}.$$

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Let $\sqrt{x-x^2}=y$.

Thus, $y\geq0$ and $$x^2-x+y^2=0$$ or $$\left(x-\frac{1}{2}\right)^2+y^2=\left(\frac{1}{2}\right)^2,$$ which is a semicircle with radius $\frac{1}{2}.$

Thus, our integral it's $$\frac{1}{2}\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{8}.$$

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  • $\begingroup$ It's a semicircle since $y>=0$ makes the domain to be only the upper semi-plane, right? $\endgroup$ – Number Dec 31 '18 at 18:56
  • $\begingroup$ Yes, of course. $\endgroup$ – Michael Rozenberg Dec 31 '18 at 18:57
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    $\begingroup$ That is awesome! I learnt something new today. $\endgroup$ – Number Dec 31 '18 at 18:58
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Another technique just for fun (and in the meanwhile, happy new year!). We have $$ \frac{1}{\sqrt{1-x}}\stackrel{L^2(0,1)}{=}2\sum_{n\geq 1}P_n(2x-1) $$ hence by Bonnet's recursion formulas and symmetry $$ \sqrt{1-x}=2\sum_{n\geq 0}\frac{1}{(1-2n)(2n+3)}P_n(2x-1) $$ $$ \sqrt{x}=2\sum_{n\geq 0}\frac{(-1)^n}{(1-2n)(2n+3)}P_n(2x-1) $$ and these FL expansions lead to $$ \int_{0}^{1}\sqrt{x(1-x)}\,dx = \color{blue}{4\sum_{n\geq 0}\frac{(-1)^n}{(1-2n)^2 (2n+1)(2n+3)^2}}. $$ Partial fraction decomposition and telescopic series convert the RHS into $$ \frac{1}{2}\sum_{n\geq 0}\frac{(-1)^n}{2n+1} = \frac{1}{2}\int_{0}^{1}\sum_{n\geq 0}(-1)^n x^{2n}\,dx = \frac{1}{2}\int_{0}^{1}\frac{dx}{1+x^2}=\frac{\pi}{8}.$$ As a by-product, we got a nice, rapidly convergent representation for $\pi$ (eight times the blue one).


Don't like the Legendre base of $L^2(0,1)$? Fine, let us go with the Chebyshev one (with respect to a different inner product). Our integral is $$ \int_{0}^{1}x(1-x)\frac{dx}{\sqrt{x(1-x)}} $$ and the orthogonality relation for the Chebyshev base is $$ \int_{0}^{1}T_m(2x-1)T_n(2x-1)\frac{dx}{\sqrt{x(1-x)}}=\frac{\pi}{2}\delta(m,n)(1+\delta(m)). $$ Since $x(1-x)$ decomposes as $\color{blue}{\frac{1}{8}}T_0(2x-1)-\frac{1}{8}T_2(2x-1)$, the value of our integral is $\frac{\pi}{8}$.

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J.G. has the elementary method down. If you see a $1-x^2$ term inside your integrand, it might be wise to give a trig substitution a try. In this case, letting $x=\sin^2\theta$ works out beautifully.

There’s another less elementary way by utilizing the beta function and the Gamma function. Let me know if you need a proof (I proudly have an elementary proof of it).

Beta Function:$$\operatorname{B}\left(m,n\right)=\int\limits_0^1\mathrm dx\,x^{m-1}(1-x)^{n-1}=\frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$

The integral under question is then simply$$\begin{align*}\mathfrak{I} & =\operatorname{B}\left(\frac 32,\frac 32\right)\\ & =\frac {1}{2}\left(\frac {\sqrt{\pi}}2\right)^2\end{align*}$$ Thus$$\int\limits_0^1\mathrm dx\, \sqrt{x(1-x)}\color{blue}{=\frac {\pi}8}$$

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  • $\begingroup$ What's your proof? $\endgroup$ – Larry Dec 31 '18 at 20:00
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    $\begingroup$ @Larry: the fact that $\int_{0}^{1}x^{1/2}(1-x)^{1/2}\,dx$ is a value of the Beta function, I guess. $\endgroup$ – Jack D'Aurizio Jan 1 at 23:36
  • $\begingroup$ Ok, Frank said he has a elementary proof. I was curious about that $\endgroup$ – Larry Jan 2 at 0:36
  • $\begingroup$ @Larry: we already have one above: $\int_{0}^{1}\sqrt{x(1-x)}\,dx$ is the area of a half-circle centered at $(1/2,0)$ through the origin. $\endgroup$ – Jack D'Aurizio Jan 2 at 1:12
  • $\begingroup$ @Larry It’s just integration by parts. $\endgroup$ – Frank W. Jan 2 at 1:23

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