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I can't calculate the following integral: $$\frac{1}{\pi} \int_a^b \frac{1}{\sqrt{x(1-x)}}dx,$$ where $[a, b] \subset [0,1]$.

Can someone, please, give me a hint?

Thank you!

g.pomegranate
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  • Hint look up Euler Substitution – MrYouMath Oct 18 '16 at 21:16
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    @MrYouMath: it is even easier than that. – Jack D'Aurizio Oct 18 '16 at 21:19
  • Or geometrically, this is the integral (which is normalized by the factor $\pi/2$) that computes the arc length of $$ (x-\tfrac{1}{2})^2 + y^2 = \tfrac{1}{4},\qquad a\leq x\leq b \text{ and } y\geq 0.$$ The you can compute the integral using trigonometry only. – Sangchul Lee Oct 18 '16 at 21:23
  • @JackD'Aurizio: I intended to show a method which is applicable to a wider range of Integrals:). – MrYouMath Oct 18 '16 at 21:42
  • @MrYouMath: I am perfectly fine with that idea, but I think it would be better to add a link to something explaining Euler's substitution, in such a case. – Jack D'Aurizio Oct 18 '16 at 21:44

3 Answers3

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If $[a,b]\subsetneq[0,1]$, by setting $x=t^2$ we get: $$ \frac{1}{\pi}\int_{a}^{b}\frac{dx}{\sqrt{x(1-x)}}=\frac{2}{\pi}\int_{\sqrt{a}}^{\sqrt{b}}\frac{dt}{\sqrt{1-t^2}}=\color{red}{\frac{2}{\pi}\left(\arcsin\sqrt{b}-\arcsin\sqrt{a}\right)}.$$

Jack D'Aurizio
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Rewrite it as. ${\displaystyle\int}\dfrac{1}{\sqrt{1-x}\sqrt{x}}\,\mathrm{d}x$. Use u-substituion with $u=\sqrt{x}, \frac{du}{dx}=\frac{1}{2\sqrt{x}}$. It then becomes $\class{steps-node}{\cssId{steps-node-1}{2}}{\displaystyle\int}\dfrac{1}{\sqrt{1-u^2}}\,\mathrm{d}u$. I'm sure you can take it from here.

AfronPie
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Anoteher simple substitution exploiting the symmetry of the integrand: $x=1/2 + u \implies dx=du$. $$\int\frac{1}{\sqrt{x(1-x)}}dx=\int\frac{1}{\sqrt{(1/2+u)(1/2-u)}}du$$

MrYouMath
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