Integral of $(1-x^2)^{1/4}$












1












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I came across this integral while originally solving the integral of $((1-sqrt{x})/(1+sqrt{x}))^{1/2}$ which led me to two integrals, one, $sqrt{x}/(1-x^2)^{1/2}$ which on substituting $(1-x^2)^{1/2}$ as t gives $2(1-t^2)^{1/4}$, which is in the title as well and the other a simple $1/(1-x^2)^{1/2}$ which yeilds $arcsin(x)$. I can't seem to find a method to solve that integral, would appreciate any help you can offer.






calculus integration







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    Hey, see my post here - math.stackexchange.com/questions/3057298/…
    $endgroup$
    – DavidG
    Dec 31 '18 at 10:56
















1












$begingroup$


I came across this integral while originally solving the integral of $((1-sqrt{x})/(1+sqrt{x}))^{1/2}$ which led me to two integrals, one, $sqrt{x}/(1-x^2)^{1/2}$ which on substituting $(1-x^2)^{1/2}$ as t gives $2(1-t^2)^{1/4}$, which is in the title as well and the other a simple $1/(1-x^2)^{1/2}$ which yeilds $arcsin(x)$. I can't seem to find a method to solve that integral, would appreciate any help you can offer.






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hey, see my post here - math.stackexchange.com/questions/3057298/…
    $endgroup$
    – DavidG
    Dec 31 '18 at 10:56














1












1








1





$begingroup$


I came across this integral while originally solving the integral of $((1-sqrt{x})/(1+sqrt{x}))^{1/2}$ which led me to two integrals, one, $sqrt{x}/(1-x^2)^{1/2}$ which on substituting $(1-x^2)^{1/2}$ as t gives $2(1-t^2)^{1/4}$, which is in the title as well and the other a simple $1/(1-x^2)^{1/2}$ which yeilds $arcsin(x)$. I can't seem to find a method to solve that integral, would appreciate any help you can offer.






calculus integration







share|cite|improve this question











$endgroup$




I came across this integral while originally solving the integral of $((1-sqrt{x})/(1+sqrt{x}))^{1/2}$ which led me to two integrals, one, $sqrt{x}/(1-x^2)^{1/2}$ which on substituting $(1-x^2)^{1/2}$ as t gives $2(1-t^2)^{1/4}$, which is in the title as well and the other a simple $1/(1-x^2)^{1/2}$ which yeilds $arcsin(x)$. I can't seem to find a method to solve that integral, would appreciate any help you can offer.






calculus integration




calculus integration






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edited Dec 30 '18 at 6:49









Fortox

6718




6718










asked Dec 30 '18 at 6:19









Vivek SinhaVivek Sinha

61




61












  • $begingroup$
    Hey, see my post here - math.stackexchange.com/questions/3057298/…
    $endgroup$
    – DavidG
    Dec 31 '18 at 10:56


















  • $begingroup$
    Hey, see my post here - math.stackexchange.com/questions/3057298/…
    $endgroup$
    – DavidG
    Dec 31 '18 at 10:56
















$begingroup$
Hey, see my post here - math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Dec 31 '18 at 10:56




$begingroup$
Hey, see my post here - math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Dec 31 '18 at 10:56










2 Answers
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This integral doesn't appear to have a elementary antiderivative---or at least neither Maple nor SageMath could find one. Maple gives
$$int (1 - x^2)^{1 / 4} dx = x cdot {}_2 F_1left(-frac{1}{4}, frac{1}{2}; frac{3}{2}; x^2right) + C,$$
where ${}_2 F_1$ is the ordinary hypergeometric function.



Incidentally, to evaluate the original integral,
$$int sqrtfrac{1 - sqrt{x}}{1 + sqrt{x}} ,dx ,$$
the substitution $x = u^2, dx = 2 u ,du$ transforms the integral into one with an integrand a rational function of $u$ and $sqrt{1 - u^2}$, which can thus in turn be rationalized with an Euler substitution.






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    2












    $begingroup$

    Note that:
    $$
    int frac{sqrt x}{sqrt{1-x^2}},dx = int (1-t^2)^{1/4}frac{-2t}{2tsqrt{1-t^2}},dt
    $$
    But in fact you can work directly with what you originally have:
    $$ I= intsqrt{frac{1-sqrt x}{1+sqrt x}},dx =int sqrt{frac{1-u}{1+u}}2u,du = int 4(z^2-1)sqrt{2-z^2} ,dz$$



    using the substitutions $x=u^2, u=z^2 -1$. Then, let $z=sqrt{2}sin{y}$ which gives you:



    $$
    I= int 4(2sin^2{y}-1)sqrt{2}cos{y}, dy = 4sqrt{2} int cos{y}sin^2{y} -cos^3{y} ,dz
    $$
    $$= frac{4sqrt{2}}{3} sin^3{y} - 4sqrt{2}(sin{y}-frac{sin^3{y}}{3}) + Constant
    $$



    Finally you can then rewrite the result in terms of the variable $x$ using:
    $$
    y=arcsin{frac{z}{sqrt{2}}}=arcsin{frac{sqrt{sqrt{x}+1}}{sqrt{2}}}
    $$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes









      2












      $begingroup$

      This integral doesn't appear to have a elementary antiderivative---or at least neither Maple nor SageMath could find one. Maple gives
      $$int (1 - x^2)^{1 / 4} dx = x cdot {}_2 F_1left(-frac{1}{4}, frac{1}{2}; frac{3}{2}; x^2right) + C,$$
      where ${}_2 F_1$ is the ordinary hypergeometric function.



      Incidentally, to evaluate the original integral,
      $$int sqrtfrac{1 - sqrt{x}}{1 + sqrt{x}} ,dx ,$$
      the substitution $x = u^2, dx = 2 u ,du$ transforms the integral into one with an integrand a rational function of $u$ and $sqrt{1 - u^2}$, which can thus in turn be rationalized with an Euler substitution.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        This integral doesn't appear to have a elementary antiderivative---or at least neither Maple nor SageMath could find one. Maple gives
        $$int (1 - x^2)^{1 / 4} dx = x cdot {}_2 F_1left(-frac{1}{4}, frac{1}{2}; frac{3}{2}; x^2right) + C,$$
        where ${}_2 F_1$ is the ordinary hypergeometric function.



        Incidentally, to evaluate the original integral,
        $$int sqrtfrac{1 - sqrt{x}}{1 + sqrt{x}} ,dx ,$$
        the substitution $x = u^2, dx = 2 u ,du$ transforms the integral into one with an integrand a rational function of $u$ and $sqrt{1 - u^2}$, which can thus in turn be rationalized with an Euler substitution.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          This integral doesn't appear to have a elementary antiderivative---or at least neither Maple nor SageMath could find one. Maple gives
          $$int (1 - x^2)^{1 / 4} dx = x cdot {}_2 F_1left(-frac{1}{4}, frac{1}{2}; frac{3}{2}; x^2right) + C,$$
          where ${}_2 F_1$ is the ordinary hypergeometric function.



          Incidentally, to evaluate the original integral,
          $$int sqrtfrac{1 - sqrt{x}}{1 + sqrt{x}} ,dx ,$$
          the substitution $x = u^2, dx = 2 u ,du$ transforms the integral into one with an integrand a rational function of $u$ and $sqrt{1 - u^2}$, which can thus in turn be rationalized with an Euler substitution.






          share|cite|improve this answer









          $endgroup$



          This integral doesn't appear to have a elementary antiderivative---or at least neither Maple nor SageMath could find one. Maple gives
          $$int (1 - x^2)^{1 / 4} dx = x cdot {}_2 F_1left(-frac{1}{4}, frac{1}{2}; frac{3}{2}; x^2right) + C,$$
          where ${}_2 F_1$ is the ordinary hypergeometric function.



          Incidentally, to evaluate the original integral,
          $$int sqrtfrac{1 - sqrt{x}}{1 + sqrt{x}} ,dx ,$$
          the substitution $x = u^2, dx = 2 u ,du$ transforms the integral into one with an integrand a rational function of $u$ and $sqrt{1 - u^2}$, which can thus in turn be rationalized with an Euler substitution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 6:34









          TravisTravis

          63.7k769151




          63.7k769151























              2












              $begingroup$

              Note that:
              $$
              int frac{sqrt x}{sqrt{1-x^2}},dx = int (1-t^2)^{1/4}frac{-2t}{2tsqrt{1-t^2}},dt
              $$
              But in fact you can work directly with what you originally have:
              $$ I= intsqrt{frac{1-sqrt x}{1+sqrt x}},dx =int sqrt{frac{1-u}{1+u}}2u,du = int 4(z^2-1)sqrt{2-z^2} ,dz$$



              using the substitutions $x=u^2, u=z^2 -1$. Then, let $z=sqrt{2}sin{y}$ which gives you:



              $$
              I= int 4(2sin^2{y}-1)sqrt{2}cos{y}, dy = 4sqrt{2} int cos{y}sin^2{y} -cos^3{y} ,dz
              $$
              $$= frac{4sqrt{2}}{3} sin^3{y} - 4sqrt{2}(sin{y}-frac{sin^3{y}}{3}) + Constant
              $$



              Finally you can then rewrite the result in terms of the variable $x$ using:
              $$
              y=arcsin{frac{z}{sqrt{2}}}=arcsin{frac{sqrt{sqrt{x}+1}}{sqrt{2}}}
              $$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Note that:
                $$
                int frac{sqrt x}{sqrt{1-x^2}},dx = int (1-t^2)^{1/4}frac{-2t}{2tsqrt{1-t^2}},dt
                $$
                But in fact you can work directly with what you originally have:
                $$ I= intsqrt{frac{1-sqrt x}{1+sqrt x}},dx =int sqrt{frac{1-u}{1+u}}2u,du = int 4(z^2-1)sqrt{2-z^2} ,dz$$



                using the substitutions $x=u^2, u=z^2 -1$. Then, let $z=sqrt{2}sin{y}$ which gives you:



                $$
                I= int 4(2sin^2{y}-1)sqrt{2}cos{y}, dy = 4sqrt{2} int cos{y}sin^2{y} -cos^3{y} ,dz
                $$
                $$= frac{4sqrt{2}}{3} sin^3{y} - 4sqrt{2}(sin{y}-frac{sin^3{y}}{3}) + Constant
                $$



                Finally you can then rewrite the result in terms of the variable $x$ using:
                $$
                y=arcsin{frac{z}{sqrt{2}}}=arcsin{frac{sqrt{sqrt{x}+1}}{sqrt{2}}}
                $$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Note that:
                  $$
                  int frac{sqrt x}{sqrt{1-x^2}},dx = int (1-t^2)^{1/4}frac{-2t}{2tsqrt{1-t^2}},dt
                  $$
                  But in fact you can work directly with what you originally have:
                  $$ I= intsqrt{frac{1-sqrt x}{1+sqrt x}},dx =int sqrt{frac{1-u}{1+u}}2u,du = int 4(z^2-1)sqrt{2-z^2} ,dz$$



                  using the substitutions $x=u^2, u=z^2 -1$. Then, let $z=sqrt{2}sin{y}$ which gives you:



                  $$
                  I= int 4(2sin^2{y}-1)sqrt{2}cos{y}, dy = 4sqrt{2} int cos{y}sin^2{y} -cos^3{y} ,dz
                  $$
                  $$= frac{4sqrt{2}}{3} sin^3{y} - 4sqrt{2}(sin{y}-frac{sin^3{y}}{3}) + Constant
                  $$



                  Finally you can then rewrite the result in terms of the variable $x$ using:
                  $$
                  y=arcsin{frac{z}{sqrt{2}}}=arcsin{frac{sqrt{sqrt{x}+1}}{sqrt{2}}}
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  Note that:
                  $$
                  int frac{sqrt x}{sqrt{1-x^2}},dx = int (1-t^2)^{1/4}frac{-2t}{2tsqrt{1-t^2}},dt
                  $$
                  But in fact you can work directly with what you originally have:
                  $$ I= intsqrt{frac{1-sqrt x}{1+sqrt x}},dx =int sqrt{frac{1-u}{1+u}}2u,du = int 4(z^2-1)sqrt{2-z^2} ,dz$$



                  using the substitutions $x=u^2, u=z^2 -1$. Then, let $z=sqrt{2}sin{y}$ which gives you:



                  $$
                  I= int 4(2sin^2{y}-1)sqrt{2}cos{y}, dy = 4sqrt{2} int cos{y}sin^2{y} -cos^3{y} ,dz
                  $$
                  $$= frac{4sqrt{2}}{3} sin^3{y} - 4sqrt{2}(sin{y}-frac{sin^3{y}}{3}) + Constant
                  $$



                  Finally you can then rewrite the result in terms of the variable $x$ using:
                  $$
                  y=arcsin{frac{z}{sqrt{2}}}=arcsin{frac{sqrt{sqrt{x}+1}}{sqrt{2}}}
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 30 '18 at 7:13









                  Test123Test123

                  2,792828




                  2,792828






























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