Is there a clever way to compute $int limits_a^b sqrt[3](b-x)(x-a)^2 dx$

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I'm wondering how to compute $int limits_a^b sqrt[3](b-x)(x-a)^2 dx$. I tried doing Chebyshe[o]v's substitution to first compute the antiderivative which led me to $int fracw^3(w^3+1)^3 dw$ what is quite unpleasant as Wolframalpha says.



Thanks.




integration radicals




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  • $$x=acos^2t+bsin^2t$$
    – lab bhattacharjee
    Jul 25 at 16:41










  • @labbhattacharjee And then what?
    – Mark Viola
    Jul 25 at 17:27














up vote
2
down vote

favorite
2












I'm wondering how to compute $int limits_a^b sqrt[3](b-x)(x-a)^2 dx$. I tried doing Chebyshe[o]v's substitution to first compute the antiderivative which led me to $int fracw^3(w^3+1)^3 dw$ what is quite unpleasant as Wolframalpha says.



Thanks.




integration radicals




share|cite|improve this question



















  • $$x=acos^2t+bsin^2t$$
    – lab bhattacharjee
    Jul 25 at 16:41










  • @labbhattacharjee And then what?
    – Mark Viola
    Jul 25 at 17:27












up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





I'm wondering how to compute $int limits_a^b sqrt[3](b-x)(x-a)^2 dx$. I tried doing Chebyshe[o]v's substitution to first compute the antiderivative which led me to $int fracw^3(w^3+1)^3 dw$ what is quite unpleasant as Wolframalpha says.



Thanks.




integration radicals




share|cite|improve this question











I'm wondering how to compute $int limits_a^b sqrt[3](b-x)(x-a)^2 dx$. I tried doing Chebyshe[o]v's substitution to first compute the antiderivative which led me to $int fracw^3(w^3+1)^3 dw$ what is quite unpleasant as Wolframalpha says.



Thanks.





integration radicals





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asked Jul 25 at 16:26









Nicholas S

17010




17010











  • $$x=acos^2t+bsin^2t$$
    – lab bhattacharjee
    Jul 25 at 16:41










  • @labbhattacharjee And then what?
    – Mark Viola
    Jul 25 at 17:27
















  • $$x=acos^2t+bsin^2t$$
    – lab bhattacharjee
    Jul 25 at 16:41










  • @labbhattacharjee And then what?
    – Mark Viola
    Jul 25 at 17:27















$$x=acos^2t+bsin^2t$$
– lab bhattacharjee
Jul 25 at 16:41




$$x=acos^2t+bsin^2t$$
– lab bhattacharjee
Jul 25 at 16:41












@labbhattacharjee And then what?
– Mark Viola
Jul 25 at 17:27




@labbhattacharjee And then what?
– Mark Viola
Jul 25 at 17:27










2 Answers
2






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2
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accepted










It's better to do this without trying to find an antiderivative. Substituting $x=a+(b-a)y$, we find that $x-a = (b-a)y$ and $b-x = (b-a)(1-y)$, so the integral reduces to
$$ (b-a)^2 int_0^1 y^2/3 (1-y)^1/3 , dy. $$
The remaining integral is $B(frac53,frac43) = 2pi/3^5/2 $ using properties of the Beta- and Gamma-functions. An elementary way of computing this Beta-function is to set $y=1/(1+u)$, which gives
$$ int_0^infty fracy^1/3(1+y)^3 , dy. $$
This can be calculated by differentiating the integral
$$ int_0^infty fracy^s-1a+y , dy = a^s-1picscpi s $$
a couple of times.






share|cite|improve this answer




























    up vote
    0
    down vote













    Thanks @Chappers for an idea. Here is the way of computing the product they was talking about:



    We will require two basic properties. The first one is the Euler reflection formula:
    $$Gamma(1-z)Gamma(z) = fracpisin z pi$$
    The second is $Gamma(z+1)=zGamma(z)$. So,
    $$Gammaleft(frac53right) Gammaleft(-frac23 right) = fracpisin left(-frac23 piright)$$
    $$Gammaleft(frac43right) Gammaleft(-frac13 right) = fracpisin left(-frac13 piright)$$
    We multiply:
    $$Gamma(5/3)Gamma(-2/3)Gamma(4/3)Gamma(-1/3) = fracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)$$
    $$Gamma(5/3)Gamma(4/3) = fracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)Gamma(-2/3)Gamma(-1/3)$$
    $$Gammaleft(frac13 right)=Gammaleft(-frac23+1 right)=-frac23Gammaleft(-frac23 right)$$ $$Gammaleft(frac23 right)=Gammaleft(-frac13+1 right)=-frac13Gammaleft(-frac13 right)$$
    Hence
    $$Gammaleft(-frac13 right) = -3Gammaleft(frac23 right) text и Gammaleft(-frac23 right) = -frac32Gammaleft(frac13 right)$$
    Finally,
    $$Gammaleft(frac53right)Gammaleft(frac43right) =displaystylefracpi^2sin left(-frac23 piright) cdot sin left(-frac13 piright)cdot -frac32 Gammaleft(frac13right) cdot(-3)Gammaleft(frac23right)=$$
    $$=frac29cdotfracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)underbraceGammaleft(1/3right) Gammaleft(2/3right)_displaystylefracpisin(1/3pi)=$$
    $$frac29cdotfracpi cdot sin (1/3pi)sin(-2/3pi)cdot sin(-1/3 pi)= frac4pi9sqrt3$$
    So:
    $$mathrmBleft(frac53,frac43 right) = fracGamma(5/3) Gamma(4/3)Gamma(3) = fracGamma(5/3) Gamma(4/3)(3-1)!=$$
    $$= fracGamma(5/3) Gamma(4/3)2=frac2pi9sqrt3$$



    In conclusion we obtain:
    $$boxeddisplaystyleint limits_a^b sqrt[3](b-x)(x-a)^2 dx = (b-a)^2frac2pi9sqrt3$$






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      2 Answers
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      2 Answers
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      up vote
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      accepted










      It's better to do this without trying to find an antiderivative. Substituting $x=a+(b-a)y$, we find that $x-a = (b-a)y$ and $b-x = (b-a)(1-y)$, so the integral reduces to
      $$ (b-a)^2 int_0^1 y^2/3 (1-y)^1/3 , dy. $$
      The remaining integral is $B(frac53,frac43) = 2pi/3^5/2 $ using properties of the Beta- and Gamma-functions. An elementary way of computing this Beta-function is to set $y=1/(1+u)$, which gives
      $$ int_0^infty fracy^1/3(1+y)^3 , dy. $$
      This can be calculated by differentiating the integral
      $$ int_0^infty fracy^s-1a+y , dy = a^s-1picscpi s $$
      a couple of times.






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted










        It's better to do this without trying to find an antiderivative. Substituting $x=a+(b-a)y$, we find that $x-a = (b-a)y$ and $b-x = (b-a)(1-y)$, so the integral reduces to
        $$ (b-a)^2 int_0^1 y^2/3 (1-y)^1/3 , dy. $$
        The remaining integral is $B(frac53,frac43) = 2pi/3^5/2 $ using properties of the Beta- and Gamma-functions. An elementary way of computing this Beta-function is to set $y=1/(1+u)$, which gives
        $$ int_0^infty fracy^1/3(1+y)^3 , dy. $$
        This can be calculated by differentiating the integral
        $$ int_0^infty fracy^s-1a+y , dy = a^s-1picscpi s $$
        a couple of times.






        share|cite|improve this answer























          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          It's better to do this without trying to find an antiderivative. Substituting $x=a+(b-a)y$, we find that $x-a = (b-a)y$ and $b-x = (b-a)(1-y)$, so the integral reduces to
          $$ (b-a)^2 int_0^1 y^2/3 (1-y)^1/3 , dy. $$
          The remaining integral is $B(frac53,frac43) = 2pi/3^5/2 $ using properties of the Beta- and Gamma-functions. An elementary way of computing this Beta-function is to set $y=1/(1+u)$, which gives
          $$ int_0^infty fracy^1/3(1+y)^3 , dy. $$
          This can be calculated by differentiating the integral
          $$ int_0^infty fracy^s-1a+y , dy = a^s-1picscpi s $$
          a couple of times.






          share|cite|improve this answer













          It's better to do this without trying to find an antiderivative. Substituting $x=a+(b-a)y$, we find that $x-a = (b-a)y$ and $b-x = (b-a)(1-y)$, so the integral reduces to
          $$ (b-a)^2 int_0^1 y^2/3 (1-y)^1/3 , dy. $$
          The remaining integral is $B(frac53,frac43) = 2pi/3^5/2 $ using properties of the Beta- and Gamma-functions. An elementary way of computing this Beta-function is to set $y=1/(1+u)$, which gives
          $$ int_0^infty fracy^1/3(1+y)^3 , dy. $$
          This can be calculated by differentiating the integral
          $$ int_0^infty fracy^s-1a+y , dy = a^s-1picscpi s $$
          a couple of times.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 25 at 16:47









          Chappers

          55k74190




          55k74190




















              up vote
              0
              down vote













              Thanks @Chappers for an idea. Here is the way of computing the product they was talking about:



              We will require two basic properties. The first one is the Euler reflection formula:
              $$Gamma(1-z)Gamma(z) = fracpisin z pi$$
              The second is $Gamma(z+1)=zGamma(z)$. So,
              $$Gammaleft(frac53right) Gammaleft(-frac23 right) = fracpisin left(-frac23 piright)$$
              $$Gammaleft(frac43right) Gammaleft(-frac13 right) = fracpisin left(-frac13 piright)$$
              We multiply:
              $$Gamma(5/3)Gamma(-2/3)Gamma(4/3)Gamma(-1/3) = fracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)$$
              $$Gamma(5/3)Gamma(4/3) = fracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)Gamma(-2/3)Gamma(-1/3)$$
              $$Gammaleft(frac13 right)=Gammaleft(-frac23+1 right)=-frac23Gammaleft(-frac23 right)$$ $$Gammaleft(frac23 right)=Gammaleft(-frac13+1 right)=-frac13Gammaleft(-frac13 right)$$
              Hence
              $$Gammaleft(-frac13 right) = -3Gammaleft(frac23 right) text и Gammaleft(-frac23 right) = -frac32Gammaleft(frac13 right)$$
              Finally,
              $$Gammaleft(frac53right)Gammaleft(frac43right) =displaystylefracpi^2sin left(-frac23 piright) cdot sin left(-frac13 piright)cdot -frac32 Gammaleft(frac13right) cdot(-3)Gammaleft(frac23right)=$$
              $$=frac29cdotfracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)underbraceGammaleft(1/3right) Gammaleft(2/3right)_displaystylefracpisin(1/3pi)=$$
              $$frac29cdotfracpi cdot sin (1/3pi)sin(-2/3pi)cdot sin(-1/3 pi)= frac4pi9sqrt3$$
              So:
              $$mathrmBleft(frac53,frac43 right) = fracGamma(5/3) Gamma(4/3)Gamma(3) = fracGamma(5/3) Gamma(4/3)(3-1)!=$$
              $$= fracGamma(5/3) Gamma(4/3)2=frac2pi9sqrt3$$



              In conclusion we obtain:
              $$boxeddisplaystyleint limits_a^b sqrt[3](b-x)(x-a)^2 dx = (b-a)^2frac2pi9sqrt3$$






              share|cite|improve this answer

























                up vote
                0
                down vote













                Thanks @Chappers for an idea. Here is the way of computing the product they was talking about:



                We will require two basic properties. The first one is the Euler reflection formula:
                $$Gamma(1-z)Gamma(z) = fracpisin z pi$$
                The second is $Gamma(z+1)=zGamma(z)$. So,
                $$Gammaleft(frac53right) Gammaleft(-frac23 right) = fracpisin left(-frac23 piright)$$
                $$Gammaleft(frac43right) Gammaleft(-frac13 right) = fracpisin left(-frac13 piright)$$
                We multiply:
                $$Gamma(5/3)Gamma(-2/3)Gamma(4/3)Gamma(-1/3) = fracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)$$
                $$Gamma(5/3)Gamma(4/3) = fracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)Gamma(-2/3)Gamma(-1/3)$$
                $$Gammaleft(frac13 right)=Gammaleft(-frac23+1 right)=-frac23Gammaleft(-frac23 right)$$ $$Gammaleft(frac23 right)=Gammaleft(-frac13+1 right)=-frac13Gammaleft(-frac13 right)$$
                Hence
                $$Gammaleft(-frac13 right) = -3Gammaleft(frac23 right) text и Gammaleft(-frac23 right) = -frac32Gammaleft(frac13 right)$$
                Finally,
                $$Gammaleft(frac53right)Gammaleft(frac43right) =displaystylefracpi^2sin left(-frac23 piright) cdot sin left(-frac13 piright)cdot -frac32 Gammaleft(frac13right) cdot(-3)Gammaleft(frac23right)=$$
                $$=frac29cdotfracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)underbraceGammaleft(1/3right) Gammaleft(2/3right)_displaystylefracpisin(1/3pi)=$$
                $$frac29cdotfracpi cdot sin (1/3pi)sin(-2/3pi)cdot sin(-1/3 pi)= frac4pi9sqrt3$$
                So:
                $$mathrmBleft(frac53,frac43 right) = fracGamma(5/3) Gamma(4/3)Gamma(3) = fracGamma(5/3) Gamma(4/3)(3-1)!=$$
                $$= fracGamma(5/3) Gamma(4/3)2=frac2pi9sqrt3$$



                In conclusion we obtain:
                $$boxeddisplaystyleint limits_a^b sqrt[3](b-x)(x-a)^2 dx = (b-a)^2frac2pi9sqrt3$$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Thanks @Chappers for an idea. Here is the way of computing the product they was talking about:



                  We will require two basic properties. The first one is the Euler reflection formula:
                  $$Gamma(1-z)Gamma(z) = fracpisin z pi$$
                  The second is $Gamma(z+1)=zGamma(z)$. So,
                  $$Gammaleft(frac53right) Gammaleft(-frac23 right) = fracpisin left(-frac23 piright)$$
                  $$Gammaleft(frac43right) Gammaleft(-frac13 right) = fracpisin left(-frac13 piright)$$
                  We multiply:
                  $$Gamma(5/3)Gamma(-2/3)Gamma(4/3)Gamma(-1/3) = fracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)$$
                  $$Gamma(5/3)Gamma(4/3) = fracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)Gamma(-2/3)Gamma(-1/3)$$
                  $$Gammaleft(frac13 right)=Gammaleft(-frac23+1 right)=-frac23Gammaleft(-frac23 right)$$ $$Gammaleft(frac23 right)=Gammaleft(-frac13+1 right)=-frac13Gammaleft(-frac13 right)$$
                  Hence
                  $$Gammaleft(-frac13 right) = -3Gammaleft(frac23 right) text и Gammaleft(-frac23 right) = -frac32Gammaleft(frac13 right)$$
                  Finally,
                  $$Gammaleft(frac53right)Gammaleft(frac43right) =displaystylefracpi^2sin left(-frac23 piright) cdot sin left(-frac13 piright)cdot -frac32 Gammaleft(frac13right) cdot(-3)Gammaleft(frac23right)=$$
                  $$=frac29cdotfracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)underbraceGammaleft(1/3right) Gammaleft(2/3right)_displaystylefracpisin(1/3pi)=$$
                  $$frac29cdotfracpi cdot sin (1/3pi)sin(-2/3pi)cdot sin(-1/3 pi)= frac4pi9sqrt3$$
                  So:
                  $$mathrmBleft(frac53,frac43 right) = fracGamma(5/3) Gamma(4/3)Gamma(3) = fracGamma(5/3) Gamma(4/3)(3-1)!=$$
                  $$= fracGamma(5/3) Gamma(4/3)2=frac2pi9sqrt3$$



                  In conclusion we obtain:
                  $$boxeddisplaystyleint limits_a^b sqrt[3](b-x)(x-a)^2 dx = (b-a)^2frac2pi9sqrt3$$






                  share|cite|improve this answer













                  Thanks @Chappers for an idea. Here is the way of computing the product they was talking about:



                  We will require two basic properties. The first one is the Euler reflection formula:
                  $$Gamma(1-z)Gamma(z) = fracpisin z pi$$
                  The second is $Gamma(z+1)=zGamma(z)$. So,
                  $$Gammaleft(frac53right) Gammaleft(-frac23 right) = fracpisin left(-frac23 piright)$$
                  $$Gammaleft(frac43right) Gammaleft(-frac13 right) = fracpisin left(-frac13 piright)$$
                  We multiply:
                  $$Gamma(5/3)Gamma(-2/3)Gamma(4/3)Gamma(-1/3) = fracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)$$
                  $$Gamma(5/3)Gamma(4/3) = fracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)Gamma(-2/3)Gamma(-1/3)$$
                  $$Gammaleft(frac13 right)=Gammaleft(-frac23+1 right)=-frac23Gammaleft(-frac23 right)$$ $$Gammaleft(frac23 right)=Gammaleft(-frac13+1 right)=-frac13Gammaleft(-frac13 right)$$
                  Hence
                  $$Gammaleft(-frac13 right) = -3Gammaleft(frac23 right) text и Gammaleft(-frac23 right) = -frac32Gammaleft(frac13 right)$$
                  Finally,
                  $$Gammaleft(frac53right)Gammaleft(frac43right) =displaystylefracpi^2sin left(-frac23 piright) cdot sin left(-frac13 piright)cdot -frac32 Gammaleft(frac13right) cdot(-3)Gammaleft(frac23right)=$$
                  $$=frac29cdotfracpi^2sin (-2/3 pi) cdot sin (-1/3 pi)underbraceGammaleft(1/3right) Gammaleft(2/3right)_displaystylefracpisin(1/3pi)=$$
                  $$frac29cdotfracpi cdot sin (1/3pi)sin(-2/3pi)cdot sin(-1/3 pi)= frac4pi9sqrt3$$
                  So:
                  $$mathrmBleft(frac53,frac43 right) = fracGamma(5/3) Gamma(4/3)Gamma(3) = fracGamma(5/3) Gamma(4/3)(3-1)!=$$
                  $$= fracGamma(5/3) Gamma(4/3)2=frac2pi9sqrt3$$



                  In conclusion we obtain:
                  $$boxeddisplaystyleint limits_a^b sqrt[3](b-x)(x-a)^2 dx = (b-a)^2frac2pi9sqrt3$$







                  share|cite|improve this answer













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                  share|cite|improve this answer











                  answered Jul 25 at 20:18









                  Nicholas S

                  17010




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