Difficult Definite Integral $int_{0}^{frac{pi}{2}} sqrt{1+2cos^2left(frac{pi}{2} - xright)} + sin x, dx$
$begingroup$
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.
calculus integration
$endgroup$
|
show 5 more comments
$begingroup$
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.
calculus integration
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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.
calculus integration
$endgroup$
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
calculus integration
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
|
show 5 more comments
$begingroup$
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
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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 $E(-2)approx 2.184438$.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
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 $E(-2)approx 2.184438$.
$endgroup$
add a comment |
$begingroup$
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 $E(-2)approx 2.184438$.
$endgroup$
add a comment |
$begingroup$
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 $E(-2)approx 2.184438$.
$endgroup$
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 $E(-2)approx 2.184438$.
edited Dec 21 '18 at 9:06
answered Dec 21 '18 at 8:01
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Dec 21 '18 at 4:03
IninterrompueIninterrompue
67519
67519
add a comment |
add a comment |
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see wolframalpha.com/input/… do you have any typo?
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– Martín Vacas Vignolo
Dec 21 '18 at 1:45
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I second @MartínVacasVignolo's comment; as written you can't really get a nice closed for answer for this.
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– DeficientMathDude
Dec 21 '18 at 1:47
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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.
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– austintice
Dec 21 '18 at 1:47
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@austintice Then I would email them to get clarification if this is a homework problem and has a due date.
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– DeficientMathDude
Dec 21 '18 at 1:48
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@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.
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– austintice
Dec 21 '18 at 1:50