How to show that $int_{0}^{1}sqrt{x}sqrt{1-x},mathrm dx =frac{pi}{8}$












2












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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.
$$int_{0}^{1}sqrt{x}sqrt{1-x},mathrm dx = left[frac{arcsin(2x-1)+sqrt{(1-x)x}(4x-2)}{8}right]_{0}^{1}=frac{pi}{8}$$






real-analysis calculus integration definite-integrals







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  • 2




    $begingroup$
    Beta function?${}$
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 18:16






  • 3




    $begingroup$
    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






  • 3




    $begingroup$
    $sqrt{x}sqrt{1-x} = sqrt{x-x^2} = sqrt{frac{1}{4}-left(x-frac{1}{2}right)^2}$; substitute $u = x-frac{1}{2}$.
    $endgroup$
    – rogerl
    Dec 31 '18 at 18:18






  • 2




    $begingroup$
    $Bleft( frac 32,frac 32right)$ ?
    $endgroup$
    – Digamma
    Jan 1 at 4:18












  • $begingroup$
    Well, solving a more general problem we can show that (when $text{n}>0$): $$mathcal{I}_text{n}:=int_0^text{n}sqrt{x}cdotsqrt{text{n}-x}spacetext{d}x=frac{pi}{8}cdottext{n}^2tag1$$
    $endgroup$
    – Jan
    Jan 1 at 16:14
















2












$begingroup$


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.
$$int_{0}^{1}sqrt{x}sqrt{1-x},mathrm dx = left[frac{arcsin(2x-1)+sqrt{(1-x)x}(4x-2)}{8}right]_{0}^{1}=frac{pi}{8}$$






real-analysis calculus integration definite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Beta function?${}$
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 18:16






  • 3




    $begingroup$
    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






  • 3




    $begingroup$
    $sqrt{x}sqrt{1-x} = sqrt{x-x^2} = sqrt{frac{1}{4}-left(x-frac{1}{2}right)^2}$; substitute $u = x-frac{1}{2}$.
    $endgroup$
    – rogerl
    Dec 31 '18 at 18:18






  • 2




    $begingroup$
    $Bleft( frac 32,frac 32right)$ ?
    $endgroup$
    – Digamma
    Jan 1 at 4:18












  • $begingroup$
    Well, solving a more general problem we can show that (when $text{n}>0$): $$mathcal{I}_text{n}:=int_0^text{n}sqrt{x}cdotsqrt{text{n}-x}spacetext{d}x=frac{pi}{8}cdottext{n}^2tag1$$
    $endgroup$
    – Jan
    Jan 1 at 16:14














2












2








2


2



$begingroup$


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.
$$int_{0}^{1}sqrt{x}sqrt{1-x},mathrm dx = left[frac{arcsin(2x-1)+sqrt{(1-x)x}(4x-2)}{8}right]_{0}^{1}=frac{pi}{8}$$






real-analysis calculus integration definite-integrals







share|cite|improve this question











$endgroup$




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.
$$int_{0}^{1}sqrt{x}sqrt{1-x},mathrm dx = left[frac{arcsin(2x-1)+sqrt{(1-x)x}(4x-2)}{8}right]_{0}^{1}=frac{pi}{8}$$






real-analysis calculus integration definite-integrals




real-analysis calculus integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 18:50









Michael Rozenberg

98.5k1590189




98.5k1590189










asked Dec 31 '18 at 18:13









LarryLarry

2,0912826




2,0912826








  • 2




    $begingroup$
    Beta function?${}$
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 18:16






  • 3




    $begingroup$
    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






  • 3




    $begingroup$
    $sqrt{x}sqrt{1-x} = sqrt{x-x^2} = sqrt{frac{1}{4}-left(x-frac{1}{2}right)^2}$; substitute $u = x-frac{1}{2}$.
    $endgroup$
    – rogerl
    Dec 31 '18 at 18:18






  • 2




    $begingroup$
    $Bleft( frac 32,frac 32right)$ ?
    $endgroup$
    – Digamma
    Jan 1 at 4:18












  • $begingroup$
    Well, solving a more general problem we can show that (when $text{n}>0$): $$mathcal{I}_text{n}:=int_0^text{n}sqrt{x}cdotsqrt{text{n}-x}spacetext{d}x=frac{pi}{8}cdottext{n}^2tag1$$
    $endgroup$
    – Jan
    Jan 1 at 16:14














  • 2




    $begingroup$
    Beta function?${}$
    $endgroup$
    – Lord Shark the Unknown
    Dec 31 '18 at 18:16






  • 3




    $begingroup$
    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






  • 3




    $begingroup$
    $sqrt{x}sqrt{1-x} = sqrt{x-x^2} = sqrt{frac{1}{4}-left(x-frac{1}{2}right)^2}$; substitute $u = x-frac{1}{2}$.
    $endgroup$
    – rogerl
    Dec 31 '18 at 18:18






  • 2




    $begingroup$
    $Bleft( frac 32,frac 32right)$ ?
    $endgroup$
    – Digamma
    Jan 1 at 4:18












  • $begingroup$
    Well, solving a more general problem we can show that (when $text{n}>0$): $$mathcal{I}_text{n}:=int_0^text{n}sqrt{x}cdotsqrt{text{n}-x}spacetext{d}x=frac{pi}{8}cdottext{n}^2tag1$$
    $endgroup$
    – Jan
    Jan 1 at 16:14








2




2




$begingroup$
Beta function?${}$
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 18:16




$begingroup$
Beta function?${}$
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 18:16




3




3




$begingroup$
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




$begingroup$
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




3




3




$begingroup$
$sqrt{x}sqrt{1-x} = sqrt{x-x^2} = sqrt{frac{1}{4}-left(x-frac{1}{2}right)^2}$; substitute $u = x-frac{1}{2}$.
$endgroup$
– rogerl
Dec 31 '18 at 18:18




$begingroup$
$sqrt{x}sqrt{1-x} = sqrt{x-x^2} = sqrt{frac{1}{4}-left(x-frac{1}{2}right)^2}$; substitute $u = x-frac{1}{2}$.
$endgroup$
– rogerl
Dec 31 '18 at 18:18




2




2




$begingroup$
$Bleft( frac 32,frac 32right)$ ?
$endgroup$
– Digamma
Jan 1 at 4:18






$begingroup$
$Bleft( frac 32,frac 32right)$ ?
$endgroup$
– Digamma
Jan 1 at 4:18














$begingroup$
Well, solving a more general problem we can show that (when $text{n}>0$): $$mathcal{I}_text{n}:=int_0^text{n}sqrt{x}cdotsqrt{text{n}-x}spacetext{d}x=frac{pi}{8}cdottext{n}^2tag1$$
$endgroup$
– Jan
Jan 1 at 16:14




$begingroup$
Well, solving a more general problem we can show that (when $text{n}>0$): $$mathcal{I}_text{n}:=int_0^text{n}sqrt{x}cdotsqrt{text{n}-x}spacetext{d}x=frac{pi}{8}cdottext{n}^2tag1$$
$endgroup$
– Jan
Jan 1 at 16:14










4 Answers
4






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18












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Write $x=sin^2 t$ so $dx=2sin tcos t dt$. Your integral becomes $$int_0^{pi/2}2sin^2 tcos^2 t dt=int_0^{pi/2}frac12 sin^2 2t dt=int_0^{pi/2}frac{1-cos 4t}{4}dt=left[frac{t}{4}-frac{1}{16}sin 4tright]_0^{pi/2}=frac{pi}{8}.$$






share|cite|improve this answer









$endgroup$





















    18












    $begingroup$

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



    Thus, $ygeq0$ 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 $$frac{1}{2}pileft(frac{1}{2}right)^2=frac{pi}{8}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It's a semicircle since $y>=0$ makes the domain to be only the upper semi-plane, right?
      $endgroup$
      – Zacky
      Dec 31 '18 at 18:56










    • $begingroup$
      Yes, of course.
      $endgroup$
      – Michael Rozenberg
      Dec 31 '18 at 18:57






    • 2




      $begingroup$
      That is awesome! I learnt something new today.
      $endgroup$
      – Zacky
      Dec 31 '18 at 18:58



















    5












    $begingroup$

    Another technique just for fun (and in the meanwhile, happy new year!). We have
    $$ frac{1}{sqrt{1-x}}stackrel{L^2(0,1)}{=}2sum_{ngeq 1}P_n(2x-1) $$
    hence by Bonnet's recursion formulas and symmetry
    $$ sqrt{1-x}=2sum_{ngeq 0}frac{1}{(1-2n)(2n+3)}P_n(2x-1) $$
    $$ sqrt{x}=2sum_{ngeq 0}frac{(-1)^n}{(1-2n)(2n+3)}P_n(2x-1) $$
    and these FL expansions lead to
    $$ int_{0}^{1}sqrt{x(1-x)},dx = color{blue}{4sum_{ngeq 0}frac{(-1)^n}{(1-2n)^2 (2n+1)(2n+3)^2}}. $$
    Partial fraction decomposition and telescopic series convert the RHS into
    $$ frac{1}{2}sum_{ngeq 0}frac{(-1)^n}{2n+1} = frac{1}{2}int_{0}^{1}sum_{ngeq 0}(-1)^n x^{2n},dx = frac{1}{2}int_{0}^{1}frac{dx}{1+x^2}=frac{pi}{8}.$$
    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 $frac{pi}{8}$.






    share|cite|improve this answer











    $endgroup$





















      4












      $begingroup$

      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 $x=sin^2theta$ 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:$$operatorname{B}left(m,nright)=intlimits_0^1mathrm dx,x^{m-1}(1-x)^{n-1}=frac {Gamma(m)Gamma(n)}{Gamma(m+n)}$$




      The integral under question is then simply$$begin{align*}mathfrak{I} & =operatorname{B}left(frac 32,frac 32right)\ & =frac {1}{2}left(frac {sqrt{pi}}2right)^2end{align*}$$
      Thus$$intlimits_0^1mathrm dx, sqrt{x(1-x)}color{blue}{=frac {pi}8}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        What's your proof?
        $endgroup$
        – Larry
        Dec 31 '18 at 20:00






      • 1




        $begingroup$
        @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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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      18












      $begingroup$

      Write $x=sin^2 t$ so $dx=2sin tcos t dt$. Your integral becomes $$int_0^{pi/2}2sin^2 tcos^2 t dt=int_0^{pi/2}frac12 sin^2 2t dt=int_0^{pi/2}frac{1-cos 4t}{4}dt=left[frac{t}{4}-frac{1}{16}sin 4tright]_0^{pi/2}=frac{pi}{8}.$$






      share|cite|improve this answer









      $endgroup$


















        18












        $begingroup$

        Write $x=sin^2 t$ so $dx=2sin tcos t dt$. Your integral becomes $$int_0^{pi/2}2sin^2 tcos^2 t dt=int_0^{pi/2}frac12 sin^2 2t dt=int_0^{pi/2}frac{1-cos 4t}{4}dt=left[frac{t}{4}-frac{1}{16}sin 4tright]_0^{pi/2}=frac{pi}{8}.$$






        share|cite|improve this answer









        $endgroup$
















          18












          18








          18





          $begingroup$

          Write $x=sin^2 t$ so $dx=2sin tcos t dt$. Your integral becomes $$int_0^{pi/2}2sin^2 tcos^2 t dt=int_0^{pi/2}frac12 sin^2 2t dt=int_0^{pi/2}frac{1-cos 4t}{4}dt=left[frac{t}{4}-frac{1}{16}sin 4tright]_0^{pi/2}=frac{pi}{8}.$$






          share|cite|improve this answer









          $endgroup$



          Write $x=sin^2 t$ so $dx=2sin tcos t dt$. Your integral becomes $$int_0^{pi/2}2sin^2 tcos^2 t dt=int_0^{pi/2}frac12 sin^2 2t dt=int_0^{pi/2}frac{1-cos 4t}{4}dt=left[frac{t}{4}-frac{1}{16}sin 4tright]_0^{pi/2}=frac{pi}{8}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 18:20









          J.G.J.G.

          24.4k22539




          24.4k22539























              18












              $begingroup$

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



              Thus, $ygeq0$ 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 $$frac{1}{2}pileft(frac{1}{2}right)^2=frac{pi}{8}.$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                It's a semicircle since $y>=0$ makes the domain to be only the upper semi-plane, right?
                $endgroup$
                – Zacky
                Dec 31 '18 at 18:56










              • $begingroup$
                Yes, of course.
                $endgroup$
                – Michael Rozenberg
                Dec 31 '18 at 18:57






              • 2




                $begingroup$
                That is awesome! I learnt something new today.
                $endgroup$
                – Zacky
                Dec 31 '18 at 18:58
















              18












              $begingroup$

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



              Thus, $ygeq0$ 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 $$frac{1}{2}pileft(frac{1}{2}right)^2=frac{pi}{8}.$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                It's a semicircle since $y>=0$ makes the domain to be only the upper semi-plane, right?
                $endgroup$
                – Zacky
                Dec 31 '18 at 18:56










              • $begingroup$
                Yes, of course.
                $endgroup$
                – Michael Rozenberg
                Dec 31 '18 at 18:57






              • 2




                $begingroup$
                That is awesome! I learnt something new today.
                $endgroup$
                – Zacky
                Dec 31 '18 at 18:58














              18












              18








              18





              $begingroup$

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



              Thus, $ygeq0$ 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 $$frac{1}{2}pileft(frac{1}{2}right)^2=frac{pi}{8}.$$






              share|cite|improve this answer









              $endgroup$



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



              Thus, $ygeq0$ 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 $$frac{1}{2}pileft(frac{1}{2}right)^2=frac{pi}{8}.$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 31 '18 at 18:45









              Michael RozenbergMichael Rozenberg

              98.5k1590189




              98.5k1590189












              • $begingroup$
                It's a semicircle since $y>=0$ makes the domain to be only the upper semi-plane, right?
                $endgroup$
                – Zacky
                Dec 31 '18 at 18:56










              • $begingroup$
                Yes, of course.
                $endgroup$
                – Michael Rozenberg
                Dec 31 '18 at 18:57






              • 2




                $begingroup$
                That is awesome! I learnt something new today.
                $endgroup$
                – Zacky
                Dec 31 '18 at 18:58


















              • $begingroup$
                It's a semicircle since $y>=0$ makes the domain to be only the upper semi-plane, right?
                $endgroup$
                – Zacky
                Dec 31 '18 at 18:56










              • $begingroup$
                Yes, of course.
                $endgroup$
                – Michael Rozenberg
                Dec 31 '18 at 18:57






              • 2




                $begingroup$
                That is awesome! I learnt something new today.
                $endgroup$
                – Zacky
                Dec 31 '18 at 18:58
















              $begingroup$
              It's a semicircle since $y>=0$ makes the domain to be only the upper semi-plane, right?
              $endgroup$
              – Zacky
              Dec 31 '18 at 18:56




              $begingroup$
              It's a semicircle since $y>=0$ makes the domain to be only the upper semi-plane, right?
              $endgroup$
              – Zacky
              Dec 31 '18 at 18:56












              $begingroup$
              Yes, of course.
              $endgroup$
              – Michael Rozenberg
              Dec 31 '18 at 18:57




              $begingroup$
              Yes, of course.
              $endgroup$
              – Michael Rozenberg
              Dec 31 '18 at 18:57




              2




              2




              $begingroup$
              That is awesome! I learnt something new today.
              $endgroup$
              – Zacky
              Dec 31 '18 at 18:58




              $begingroup$
              That is awesome! I learnt something new today.
              $endgroup$
              – Zacky
              Dec 31 '18 at 18:58











              5












              $begingroup$

              Another technique just for fun (and in the meanwhile, happy new year!). We have
              $$ frac{1}{sqrt{1-x}}stackrel{L^2(0,1)}{=}2sum_{ngeq 1}P_n(2x-1) $$
              hence by Bonnet's recursion formulas and symmetry
              $$ sqrt{1-x}=2sum_{ngeq 0}frac{1}{(1-2n)(2n+3)}P_n(2x-1) $$
              $$ sqrt{x}=2sum_{ngeq 0}frac{(-1)^n}{(1-2n)(2n+3)}P_n(2x-1) $$
              and these FL expansions lead to
              $$ int_{0}^{1}sqrt{x(1-x)},dx = color{blue}{4sum_{ngeq 0}frac{(-1)^n}{(1-2n)^2 (2n+1)(2n+3)^2}}. $$
              Partial fraction decomposition and telescopic series convert the RHS into
              $$ frac{1}{2}sum_{ngeq 0}frac{(-1)^n}{2n+1} = frac{1}{2}int_{0}^{1}sum_{ngeq 0}(-1)^n x^{2n},dx = frac{1}{2}int_{0}^{1}frac{dx}{1+x^2}=frac{pi}{8}.$$
              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 $frac{pi}{8}$.






              share|cite|improve this answer











              $endgroup$


















                5












                $begingroup$

                Another technique just for fun (and in the meanwhile, happy new year!). We have
                $$ frac{1}{sqrt{1-x}}stackrel{L^2(0,1)}{=}2sum_{ngeq 1}P_n(2x-1) $$
                hence by Bonnet's recursion formulas and symmetry
                $$ sqrt{1-x}=2sum_{ngeq 0}frac{1}{(1-2n)(2n+3)}P_n(2x-1) $$
                $$ sqrt{x}=2sum_{ngeq 0}frac{(-1)^n}{(1-2n)(2n+3)}P_n(2x-1) $$
                and these FL expansions lead to
                $$ int_{0}^{1}sqrt{x(1-x)},dx = color{blue}{4sum_{ngeq 0}frac{(-1)^n}{(1-2n)^2 (2n+1)(2n+3)^2}}. $$
                Partial fraction decomposition and telescopic series convert the RHS into
                $$ frac{1}{2}sum_{ngeq 0}frac{(-1)^n}{2n+1} = frac{1}{2}int_{0}^{1}sum_{ngeq 0}(-1)^n x^{2n},dx = frac{1}{2}int_{0}^{1}frac{dx}{1+x^2}=frac{pi}{8}.$$
                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 $frac{pi}{8}$.






                share|cite|improve this answer











                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  Another technique just for fun (and in the meanwhile, happy new year!). We have
                  $$ frac{1}{sqrt{1-x}}stackrel{L^2(0,1)}{=}2sum_{ngeq 1}P_n(2x-1) $$
                  hence by Bonnet's recursion formulas and symmetry
                  $$ sqrt{1-x}=2sum_{ngeq 0}frac{1}{(1-2n)(2n+3)}P_n(2x-1) $$
                  $$ sqrt{x}=2sum_{ngeq 0}frac{(-1)^n}{(1-2n)(2n+3)}P_n(2x-1) $$
                  and these FL expansions lead to
                  $$ int_{0}^{1}sqrt{x(1-x)},dx = color{blue}{4sum_{ngeq 0}frac{(-1)^n}{(1-2n)^2 (2n+1)(2n+3)^2}}. $$
                  Partial fraction decomposition and telescopic series convert the RHS into
                  $$ frac{1}{2}sum_{ngeq 0}frac{(-1)^n}{2n+1} = frac{1}{2}int_{0}^{1}sum_{ngeq 0}(-1)^n x^{2n},dx = frac{1}{2}int_{0}^{1}frac{dx}{1+x^2}=frac{pi}{8}.$$
                  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 $frac{pi}{8}$.






                  share|cite|improve this answer











                  $endgroup$



                  Another technique just for fun (and in the meanwhile, happy new year!). We have
                  $$ frac{1}{sqrt{1-x}}stackrel{L^2(0,1)}{=}2sum_{ngeq 1}P_n(2x-1) $$
                  hence by Bonnet's recursion formulas and symmetry
                  $$ sqrt{1-x}=2sum_{ngeq 0}frac{1}{(1-2n)(2n+3)}P_n(2x-1) $$
                  $$ sqrt{x}=2sum_{ngeq 0}frac{(-1)^n}{(1-2n)(2n+3)}P_n(2x-1) $$
                  and these FL expansions lead to
                  $$ int_{0}^{1}sqrt{x(1-x)},dx = color{blue}{4sum_{ngeq 0}frac{(-1)^n}{(1-2n)^2 (2n+1)(2n+3)^2}}. $$
                  Partial fraction decomposition and telescopic series convert the RHS into
                  $$ frac{1}{2}sum_{ngeq 0}frac{(-1)^n}{2n+1} = frac{1}{2}int_{0}^{1}sum_{ngeq 0}(-1)^n x^{2n},dx = frac{1}{2}int_{0}^{1}frac{dx}{1+x^2}=frac{pi}{8}.$$
                  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 $frac{pi}{8}$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 2 at 1:00

























                  answered Dec 31 '18 at 23:54









                  Jack D'AurizioJack D'Aurizio

                  288k33280660




                  288k33280660























                      4












                      $begingroup$

                      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 $x=sin^2theta$ 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:$$operatorname{B}left(m,nright)=intlimits_0^1mathrm dx,x^{m-1}(1-x)^{n-1}=frac {Gamma(m)Gamma(n)}{Gamma(m+n)}$$




                      The integral under question is then simply$$begin{align*}mathfrak{I} & =operatorname{B}left(frac 32,frac 32right)\ & =frac {1}{2}left(frac {sqrt{pi}}2right)^2end{align*}$$
                      Thus$$intlimits_0^1mathrm dx, sqrt{x(1-x)}color{blue}{=frac {pi}8}$$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        What's your proof?
                        $endgroup$
                        – Larry
                        Dec 31 '18 at 20:00






                      • 1




                        $begingroup$
                        @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
















                      4












                      $begingroup$

                      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 $x=sin^2theta$ 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:$$operatorname{B}left(m,nright)=intlimits_0^1mathrm dx,x^{m-1}(1-x)^{n-1}=frac {Gamma(m)Gamma(n)}{Gamma(m+n)}$$




                      The integral under question is then simply$$begin{align*}mathfrak{I} & =operatorname{B}left(frac 32,frac 32right)\ & =frac {1}{2}left(frac {sqrt{pi}}2right)^2end{align*}$$
                      Thus$$intlimits_0^1mathrm dx, sqrt{x(1-x)}color{blue}{=frac {pi}8}$$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        What's your proof?
                        $endgroup$
                        – Larry
                        Dec 31 '18 at 20:00






                      • 1




                        $begingroup$
                        @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














                      4












                      4








                      4





                      $begingroup$

                      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 $x=sin^2theta$ 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:$$operatorname{B}left(m,nright)=intlimits_0^1mathrm dx,x^{m-1}(1-x)^{n-1}=frac {Gamma(m)Gamma(n)}{Gamma(m+n)}$$




                      The integral under question is then simply$$begin{align*}mathfrak{I} & =operatorname{B}left(frac 32,frac 32right)\ & =frac {1}{2}left(frac {sqrt{pi}}2right)^2end{align*}$$
                      Thus$$intlimits_0^1mathrm dx, sqrt{x(1-x)}color{blue}{=frac {pi}8}$$






                      share|cite|improve this answer











                      $endgroup$



                      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 $x=sin^2theta$ 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:$$operatorname{B}left(m,nright)=intlimits_0^1mathrm dx,x^{m-1}(1-x)^{n-1}=frac {Gamma(m)Gamma(n)}{Gamma(m+n)}$$




                      The integral under question is then simply$$begin{align*}mathfrak{I} & =operatorname{B}left(frac 32,frac 32right)\ & =frac {1}{2}left(frac {sqrt{pi}}2right)^2end{align*}$$
                      Thus$$intlimits_0^1mathrm dx, sqrt{x(1-x)}color{blue}{=frac {pi}8}$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 31 '18 at 18:49

























                      answered Dec 31 '18 at 18:40









                      Frank W.Frank W.

                      3,3941321




                      3,3941321












                      • $begingroup$
                        What's your proof?
                        $endgroup$
                        – Larry
                        Dec 31 '18 at 20:00






                      • 1




                        $begingroup$
                        @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


















                      • $begingroup$
                        What's your proof?
                        $endgroup$
                        – Larry
                        Dec 31 '18 at 20:00






                      • 1




                        $begingroup$
                        @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
















                      $begingroup$
                      What's your proof?
                      $endgroup$
                      – Larry
                      Dec 31 '18 at 20:00




                      $begingroup$
                      What's your proof?
                      $endgroup$
                      – Larry
                      Dec 31 '18 at 20:00




                      1




                      1




                      $begingroup$
                      @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




                      $begingroup$
                      @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




                      $begingroup$
                      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: 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




                      $begingroup$
                      @Larry It’s just integration by parts.
                      $endgroup$
                      – Frank W.
                      Jan 2 at 1:23


















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