Difficult Definite Integral $int_{0}^{frac{pi}{2}} sqrt{1+2cos^2left(frac{pi}{2} - xright)} + sin x, dx$












10












$begingroup$


I have spent several days trying to solve this integral, but to no avail. This isn't from a textbook, but a challenge problem given to me by a professor. I am not looking for anyone to give me the solution, but just to lead me in the right direction.



The problem is to compute the following integral:



begin{equation}
int_{0}^{frac{pi}{2}} sqrt{1+2cos^2left(frac{pi}{2} - xright)} + sin x, dx
end{equation}



When first approaching this problem I tried to utilize the cofunction identity:
begin{equation}
cosleft(frac{pi}{2}-xright) = sin x
end{equation}



The integral then became:
begin{equation}
int_{0}^{frac{pi}{2}} sqrt{1+2sin^2x} + sin x, dx
end{equation}



I have tried several things from this point such as using the formulas
begin{equation}
sin^2x = frac{1}{2}[1-cos(2x)]
end{equation}



The integral then became:



begin{equation}
int_{0}^{frac{pi}{2}} sqrt{2-cos(2x)} + sin x, dx
end{equation}



The issue is I have tried several run arounds(of which I will not post each) with identities and other methods, but I seem to be hitting dead ends. Also, I want to mention that I'm trying to solve this using elementary methods only. I only have experience up to calculus II. Any constructive criticism or comments would be greatly appreciated! Thank you.










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$endgroup$












  • $begingroup$
    see wolframalpha.com/input/… do you have any typo?
    $endgroup$
    – Martín Vacas Vignolo
    Dec 21 '18 at 1:45










  • $begingroup$
    I second @MartínVacasVignolo's comment; as written you can't really get a nice closed for answer for this.
    $endgroup$
    – DeficientMathDude
    Dec 21 '18 at 1:47










  • $begingroup$
    No typo as far as I'm aware. The person who posed the problem may have meant to put something else, but this is the integral I was given.
    $endgroup$
    – austintice
    Dec 21 '18 at 1:47










  • $begingroup$
    @austintice Then I would email them to get clarification if this is a homework problem and has a due date.
    $endgroup$
    – DeficientMathDude
    Dec 21 '18 at 1:48








  • 1




    $begingroup$
    @DeficientMathDude like I stated this isn't a homework problem, but merely a challenge integral problem given to me by one of my research professors to take a shot at over winter break. I have given this problem ~week of time, but just keep getting looped around. I did look up and see the elliptic integral on wikipedia, but was not aware of this before.
    $endgroup$
    – austintice
    Dec 21 '18 at 1:50
















10












$begingroup$


I have spent several days trying to solve this integral, but to no avail. This isn't from a textbook, but a challenge problem given to me by a professor. I am not looking for anyone to give me the solution, but just to lead me in the right direction.



The problem is to compute the following integral:



begin{equation}
int_{0}^{frac{pi}{2}} sqrt{1+2cos^2left(frac{pi}{2} - xright)} + sin x, dx
end{equation}



When first approaching this problem I tried to utilize the cofunction identity:
begin{equation}
cosleft(frac{pi}{2}-xright) = sin x
end{equation}



The integral then became:
begin{equation}
int_{0}^{frac{pi}{2}} sqrt{1+2sin^2x} + sin x, dx
end{equation}



I have tried several things from this point such as using the formulas
begin{equation}
sin^2x = frac{1}{2}[1-cos(2x)]
end{equation}



The integral then became:



begin{equation}
int_{0}^{frac{pi}{2}} sqrt{2-cos(2x)} + sin x, dx
end{equation}



The issue is I have tried several run arounds(of which I will not post each) with identities and other methods, but I seem to be hitting dead ends. Also, I want to mention that I'm trying to solve this using elementary methods only. I only have experience up to calculus II. Any constructive criticism or comments would be greatly appreciated! Thank you.










share|cite|improve this question











$endgroup$












  • $begingroup$
    see wolframalpha.com/input/… do you have any typo?
    $endgroup$
    – Martín Vacas Vignolo
    Dec 21 '18 at 1:45










  • $begingroup$
    I second @MartínVacasVignolo's comment; as written you can't really get a nice closed for answer for this.
    $endgroup$
    – DeficientMathDude
    Dec 21 '18 at 1:47










  • $begingroup$
    No typo as far as I'm aware. The person who posed the problem may have meant to put something else, but this is the integral I was given.
    $endgroup$
    – austintice
    Dec 21 '18 at 1:47










  • $begingroup$
    @austintice Then I would email them to get clarification if this is a homework problem and has a due date.
    $endgroup$
    – DeficientMathDude
    Dec 21 '18 at 1:48








  • 1




    $begingroup$
    @DeficientMathDude like I stated this isn't a homework problem, but merely a challenge integral problem given to me by one of my research professors to take a shot at over winter break. I have given this problem ~week of time, but just keep getting looped around. I did look up and see the elliptic integral on wikipedia, but was not aware of this before.
    $endgroup$
    – austintice
    Dec 21 '18 at 1:50














10












10








10


2



$begingroup$


I have spent several days trying to solve this integral, but to no avail. This isn't from a textbook, but a challenge problem given to me by a professor. I am not looking for anyone to give me the solution, but just to lead me in the right direction.



The problem is to compute the following integral:



begin{equation}
int_{0}^{frac{pi}{2}} sqrt{1+2cos^2left(frac{pi}{2} - xright)} + sin x, dx
end{equation}



When first approaching this problem I tried to utilize the cofunction identity:
begin{equation}
cosleft(frac{pi}{2}-xright) = sin x
end{equation}



The integral then became:
begin{equation}
int_{0}^{frac{pi}{2}} sqrt{1+2sin^2x} + sin x, dx
end{equation}



I have tried several things from this point such as using the formulas
begin{equation}
sin^2x = frac{1}{2}[1-cos(2x)]
end{equation}



The integral then became:



begin{equation}
int_{0}^{frac{pi}{2}} sqrt{2-cos(2x)} + sin x, dx
end{equation}



The issue is I have tried several run arounds(of which I will not post each) with identities and other methods, but I seem to be hitting dead ends. Also, I want to mention that I'm trying to solve this using elementary methods only. I only have experience up to calculus II. Any constructive criticism or comments would be greatly appreciated! Thank you.










share|cite|improve this question











$endgroup$




I have spent several days trying to solve this integral, but to no avail. This isn't from a textbook, but a challenge problem given to me by a professor. I am not looking for anyone to give me the solution, but just to lead me in the right direction.



The problem is to compute the following integral:



begin{equation}
int_{0}^{frac{pi}{2}} sqrt{1+2cos^2left(frac{pi}{2} - xright)} + sin x, dx
end{equation}



When first approaching this problem I tried to utilize the cofunction identity:
begin{equation}
cosleft(frac{pi}{2}-xright) = sin x
end{equation}



The integral then became:
begin{equation}
int_{0}^{frac{pi}{2}} sqrt{1+2sin^2x} + sin x, dx
end{equation}



I have tried several things from this point such as using the formulas
begin{equation}
sin^2x = frac{1}{2}[1-cos(2x)]
end{equation}



The integral then became:



begin{equation}
int_{0}^{frac{pi}{2}} sqrt{2-cos(2x)} + sin x, dx
end{equation}



The issue is I have tried several run arounds(of which I will not post each) with identities and other methods, but I seem to be hitting dead ends. Also, I want to mention that I'm trying to solve this using elementary methods only. I only have experience up to calculus II. Any constructive criticism or comments would be greatly appreciated! Thank you.







calculus integration






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 1:58









Kemono Chen

3,1991844




3,1991844










asked Dec 21 '18 at 1:37









austinticeaustintice

514




514












  • $begingroup$
    see wolframalpha.com/input/… do you have any typo?
    $endgroup$
    – Martín Vacas Vignolo
    Dec 21 '18 at 1:45










  • $begingroup$
    I second @MartínVacasVignolo's comment; as written you can't really get a nice closed for answer for this.
    $endgroup$
    – DeficientMathDude
    Dec 21 '18 at 1:47










  • $begingroup$
    No typo as far as I'm aware. The person who posed the problem may have meant to put something else, but this is the integral I was given.
    $endgroup$
    – austintice
    Dec 21 '18 at 1:47










  • $begingroup$
    @austintice Then I would email them to get clarification if this is a homework problem and has a due date.
    $endgroup$
    – DeficientMathDude
    Dec 21 '18 at 1:48








  • 1




    $begingroup$
    @DeficientMathDude like I stated this isn't a homework problem, but merely a challenge integral problem given to me by one of my research professors to take a shot at over winter break. I have given this problem ~week of time, but just keep getting looped around. I did look up and see the elliptic integral on wikipedia, but was not aware of this before.
    $endgroup$
    – austintice
    Dec 21 '18 at 1:50


















  • $begingroup$
    see wolframalpha.com/input/… do you have any typo?
    $endgroup$
    – Martín Vacas Vignolo
    Dec 21 '18 at 1:45










  • $begingroup$
    I second @MartínVacasVignolo's comment; as written you can't really get a nice closed for answer for this.
    $endgroup$
    – DeficientMathDude
    Dec 21 '18 at 1:47










  • $begingroup$
    No typo as far as I'm aware. The person who posed the problem may have meant to put something else, but this is the integral I was given.
    $endgroup$
    – austintice
    Dec 21 '18 at 1:47










  • $begingroup$
    @austintice Then I would email them to get clarification if this is a homework problem and has a due date.
    $endgroup$
    – DeficientMathDude
    Dec 21 '18 at 1:48








  • 1




    $begingroup$
    @DeficientMathDude like I stated this isn't a homework problem, but merely a challenge integral problem given to me by one of my research professors to take a shot at over winter break. I have given this problem ~week of time, but just keep getting looped around. I did look up and see the elliptic integral on wikipedia, but was not aware of this before.
    $endgroup$
    – austintice
    Dec 21 '18 at 1:50
















$begingroup$
see wolframalpha.com/input/… do you have any typo?
$endgroup$
– Martín Vacas Vignolo
Dec 21 '18 at 1:45




$begingroup$
see wolframalpha.com/input/… do you have any typo?
$endgroup$
– Martín Vacas Vignolo
Dec 21 '18 at 1:45












$begingroup$
I second @MartínVacasVignolo's comment; as written you can't really get a nice closed for answer for this.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 1:47




$begingroup$
I second @MartínVacasVignolo's comment; as written you can't really get a nice closed for answer for this.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 1:47












$begingroup$
No typo as far as I'm aware. The person who posed the problem may have meant to put something else, but this is the integral I was given.
$endgroup$
– austintice
Dec 21 '18 at 1:47




$begingroup$
No typo as far as I'm aware. The person who posed the problem may have meant to put something else, but this is the integral I was given.
$endgroup$
– austintice
Dec 21 '18 at 1:47












$begingroup$
@austintice Then I would email them to get clarification if this is a homework problem and has a due date.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 1:48






$begingroup$
@austintice Then I would email them to get clarification if this is a homework problem and has a due date.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 1:48






1




1




$begingroup$
@DeficientMathDude like I stated this isn't a homework problem, but merely a challenge integral problem given to me by one of my research professors to take a shot at over winter break. I have given this problem ~week of time, but just keep getting looped around. I did look up and see the elliptic integral on wikipedia, but was not aware of this before.
$endgroup$
– austintice
Dec 21 '18 at 1:50




$begingroup$
@DeficientMathDude like I stated this isn't a homework problem, but merely a challenge integral problem given to me by one of my research professors to take a shot at over winter break. I have given this problem ~week of time, but just keep getting looped around. I did look up and see the elliptic integral on wikipedia, but was not aware of this before.
$endgroup$
– austintice
Dec 21 '18 at 1:50










2 Answers
2






active

oldest

votes


















5












$begingroup$

As said in comments and answers, you are facing elliptic integrals that you cannot evaluate easily.
$$int_0^{frac pi 2}sqrt{1+k sin ^2(x)},dx=E(-k)$$ where appears the complete elliptic integral of the second kind.



However, we can build quite good approximations. I give you one I produced years ago (for rather small values of $k$) using Padé approximants built at $k=0$.



$$E(-k) simeq frac pi 2 ,frac{1+frac{39575 }{28464}k+frac{20621} {37952}k^2+frac{129235}{2428928}k^3 } {1+frac{32459}{28464}k+frac{34741 }{113856}k^2+frac{79037
}{7286784}k ^3 }$$
which is quite good for the range $0leq k leq 4$.



Using $k=-2$, we should get , as an approximation, $frac{5810969}{8357946}piapprox 2.18423$ while the exact value would be $E(-2)approx 2.18444$.



Edit



The approximation I wrote was made more than fourty years ago and it was, at that time, a hard work. Just for the fun of it, I made, after answering, a better one which took me a few minutes .... thanks to a CAS. It is
$$E(-k) simeq frac pi 2 ,frac{1+frac{133542997 }{70902928}k+frac{1325913585 }{1134446848}k^2+frac{1210596065
}{4537787392}k^3+frac{4808786003 }{290418393088} k^4} {1+frac{115817265 }{70902928}k+frac{915821721 }{1134446848}k^2+frac{553597479
}{4537787392}k^3+frac{777708891 }{290418393088}k^4 }$$

Using $k=-2$, we should get , as an approximation, $frac{214931493555 }{309110015222}piapprox 2.184424$ while the exact value would be $E(-2)approx 2.184438$.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    You cannot solve the integral using elementary methods. It can be written in terms of a special function called the elliptic integral of the second kind $E(m)$, defined as



    $$ E(m) = int_{0}^{pi/2}sqrt{1-msin^{2}x},mathrm{d}x. $$



    This function has a power series, but that series is also hard to derive without using other special functions.






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      As said in comments and answers, you are facing elliptic integrals that you cannot evaluate easily.
      $$int_0^{frac pi 2}sqrt{1+k sin ^2(x)},dx=E(-k)$$ where appears the complete elliptic integral of the second kind.



      However, we can build quite good approximations. I give you one I produced years ago (for rather small values of $k$) using Padé approximants built at $k=0$.



      $$E(-k) simeq frac pi 2 ,frac{1+frac{39575 }{28464}k+frac{20621} {37952}k^2+frac{129235}{2428928}k^3 } {1+frac{32459}{28464}k+frac{34741 }{113856}k^2+frac{79037
      }{7286784}k ^3 }$$
      which is quite good for the range $0leq k leq 4$.



      Using $k=-2$, we should get , as an approximation, $frac{5810969}{8357946}piapprox 2.18423$ while the exact value would be $E(-2)approx 2.18444$.



      Edit



      The approximation I wrote was made more than fourty years ago and it was, at that time, a hard work. Just for the fun of it, I made, after answering, a better one which took me a few minutes .... thanks to a CAS. It is
      $$E(-k) simeq frac pi 2 ,frac{1+frac{133542997 }{70902928}k+frac{1325913585 }{1134446848}k^2+frac{1210596065
      }{4537787392}k^3+frac{4808786003 }{290418393088} k^4} {1+frac{115817265 }{70902928}k+frac{915821721 }{1134446848}k^2+frac{553597479
      }{4537787392}k^3+frac{777708891 }{290418393088}k^4 }$$

      Using $k=-2$, we should get , as an approximation, $frac{214931493555 }{309110015222}piapprox 2.184424$ while the exact value would be $E(-2)approx 2.184438$.






      share|cite|improve this answer











      $endgroup$


















        5












        $begingroup$

        As said in comments and answers, you are facing elliptic integrals that you cannot evaluate easily.
        $$int_0^{frac pi 2}sqrt{1+k sin ^2(x)},dx=E(-k)$$ where appears the complete elliptic integral of the second kind.



        However, we can build quite good approximations. I give you one I produced years ago (for rather small values of $k$) using Padé approximants built at $k=0$.



        $$E(-k) simeq frac pi 2 ,frac{1+frac{39575 }{28464}k+frac{20621} {37952}k^2+frac{129235}{2428928}k^3 } {1+frac{32459}{28464}k+frac{34741 }{113856}k^2+frac{79037
        }{7286784}k ^3 }$$
        which is quite good for the range $0leq k leq 4$.



        Using $k=-2$, we should get , as an approximation, $frac{5810969}{8357946}piapprox 2.18423$ while the exact value would be $E(-2)approx 2.18444$.



        Edit



        The approximation I wrote was made more than fourty years ago and it was, at that time, a hard work. Just for the fun of it, I made, after answering, a better one which took me a few minutes .... thanks to a CAS. It is
        $$E(-k) simeq frac pi 2 ,frac{1+frac{133542997 }{70902928}k+frac{1325913585 }{1134446848}k^2+frac{1210596065
        }{4537787392}k^3+frac{4808786003 }{290418393088} k^4} {1+frac{115817265 }{70902928}k+frac{915821721 }{1134446848}k^2+frac{553597479
        }{4537787392}k^3+frac{777708891 }{290418393088}k^4 }$$

        Using $k=-2$, we should get , as an approximation, $frac{214931493555 }{309110015222}piapprox 2.184424$ while the exact value would be $E(-2)approx 2.184438$.






        share|cite|improve this answer











        $endgroup$
















          5












          5








          5





          $begingroup$

          As said in comments and answers, you are facing elliptic integrals that you cannot evaluate easily.
          $$int_0^{frac pi 2}sqrt{1+k sin ^2(x)},dx=E(-k)$$ where appears the complete elliptic integral of the second kind.



          However, we can build quite good approximations. I give you one I produced years ago (for rather small values of $k$) using Padé approximants built at $k=0$.



          $$E(-k) simeq frac pi 2 ,frac{1+frac{39575 }{28464}k+frac{20621} {37952}k^2+frac{129235}{2428928}k^3 } {1+frac{32459}{28464}k+frac{34741 }{113856}k^2+frac{79037
          }{7286784}k ^3 }$$
          which is quite good for the range $0leq k leq 4$.



          Using $k=-2$, we should get , as an approximation, $frac{5810969}{8357946}piapprox 2.18423$ while the exact value would be $E(-2)approx 2.18444$.



          Edit



          The approximation I wrote was made more than fourty years ago and it was, at that time, a hard work. Just for the fun of it, I made, after answering, a better one which took me a few minutes .... thanks to a CAS. It is
          $$E(-k) simeq frac pi 2 ,frac{1+frac{133542997 }{70902928}k+frac{1325913585 }{1134446848}k^2+frac{1210596065
          }{4537787392}k^3+frac{4808786003 }{290418393088} k^4} {1+frac{115817265 }{70902928}k+frac{915821721 }{1134446848}k^2+frac{553597479
          }{4537787392}k^3+frac{777708891 }{290418393088}k^4 }$$

          Using $k=-2$, we should get , as an approximation, $frac{214931493555 }{309110015222}piapprox 2.184424$ while the exact value would be $E(-2)approx 2.184438$.






          share|cite|improve this answer











          $endgroup$



          As said in comments and answers, you are facing elliptic integrals that you cannot evaluate easily.
          $$int_0^{frac pi 2}sqrt{1+k sin ^2(x)},dx=E(-k)$$ where appears the complete elliptic integral of the second kind.



          However, we can build quite good approximations. I give you one I produced years ago (for rather small values of $k$) using Padé approximants built at $k=0$.



          $$E(-k) simeq frac pi 2 ,frac{1+frac{39575 }{28464}k+frac{20621} {37952}k^2+frac{129235}{2428928}k^3 } {1+frac{32459}{28464}k+frac{34741 }{113856}k^2+frac{79037
          }{7286784}k ^3 }$$
          which is quite good for the range $0leq k leq 4$.



          Using $k=-2$, we should get , as an approximation, $frac{5810969}{8357946}piapprox 2.18423$ while the exact value would be $E(-2)approx 2.18444$.



          Edit



          The approximation I wrote was made more than fourty years ago and it was, at that time, a hard work. Just for the fun of it, I made, after answering, a better one which took me a few minutes .... thanks to a CAS. It is
          $$E(-k) simeq frac pi 2 ,frac{1+frac{133542997 }{70902928}k+frac{1325913585 }{1134446848}k^2+frac{1210596065
          }{4537787392}k^3+frac{4808786003 }{290418393088} k^4} {1+frac{115817265 }{70902928}k+frac{915821721 }{1134446848}k^2+frac{553597479
          }{4537787392}k^3+frac{777708891 }{290418393088}k^4 }$$

          Using $k=-2$, we should get , as an approximation, $frac{214931493555 }{309110015222}piapprox 2.184424$ while the exact value would be $E(-2)approx 2.184438$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 21 '18 at 9:06

























          answered Dec 21 '18 at 8:01









          Claude LeiboviciClaude Leibovici

          125k1158135




          125k1158135























              4












              $begingroup$

              You cannot solve the integral using elementary methods. It can be written in terms of a special function called the elliptic integral of the second kind $E(m)$, defined as



              $$ E(m) = int_{0}^{pi/2}sqrt{1-msin^{2}x},mathrm{d}x. $$



              This function has a power series, but that series is also hard to derive without using other special functions.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                You cannot solve the integral using elementary methods. It can be written in terms of a special function called the elliptic integral of the second kind $E(m)$, defined as



                $$ E(m) = int_{0}^{pi/2}sqrt{1-msin^{2}x},mathrm{d}x. $$



                This function has a power series, but that series is also hard to derive without using other special functions.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  You cannot solve the integral using elementary methods. It can be written in terms of a special function called the elliptic integral of the second kind $E(m)$, defined as



                  $$ E(m) = int_{0}^{pi/2}sqrt{1-msin^{2}x},mathrm{d}x. $$



                  This function has a power series, but that series is also hard to derive without using other special functions.






                  share|cite|improve this answer









                  $endgroup$



                  You cannot solve the integral using elementary methods. It can be written in terms of a special function called the elliptic integral of the second kind $E(m)$, defined as



                  $$ E(m) = int_{0}^{pi/2}sqrt{1-msin^{2}x},mathrm{d}x. $$



                  This function has a power series, but that series is also hard to derive without using other special functions.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '18 at 4:03









                  IninterrompueIninterrompue

                  67519




                  67519






























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