Evaluating the integral $int_0^{pi/4}sqrt{1-16sin^2(x)}mathop{}!mathrm dx$












4












$begingroup$


How can we evaluate this integral?
$$int_limits{0}^{pi/4}sqrt{1-16sin^2(x)}mathop{}!mathrm dx$$
I tried a substitution
$$u=4sin x,quad mathrm dx=frac{mathrm du}{sqrt{16-u^2}}$$
hence the integration will be



$$int_limits{u=0}^{u=2sqrt{2}}frac{sqrt{1-u^2}}{sqrt{16-u^2}}mathop{}!mathrm du$$
But I could not complete the solution using this substitution.






calculus integration trigonometry definite-integrals elliptic-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Accordingly, this cannot be solved in terms of elementary functions. In other words, there is no closed form solution and you must resort to numerical integration over a given interval.
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 20:08








  • 3




    $begingroup$
    You have an elliptic integral. Furthermore, at $x = frac {pi}{4}$ the integrand is complex.
    $endgroup$
    – Doug M
    Dec 13 '18 at 20:22












  • $begingroup$
    u = 2.sin x is a nice substitution but results in complex numbers.
    $endgroup$
    – William Elliot
    Dec 13 '18 at 20:30










  • $begingroup$
    You're taking the square root of negative numbers in the integral. Is that what you want?
    $endgroup$
    – zhw.
    Dec 13 '18 at 21:44
















4












$begingroup$


How can we evaluate this integral?
$$int_limits{0}^{pi/4}sqrt{1-16sin^2(x)}mathop{}!mathrm dx$$
I tried a substitution
$$u=4sin x,quad mathrm dx=frac{mathrm du}{sqrt{16-u^2}}$$
hence the integration will be



$$int_limits{u=0}^{u=2sqrt{2}}frac{sqrt{1-u^2}}{sqrt{16-u^2}}mathop{}!mathrm du$$
But I could not complete the solution using this substitution.






calculus integration trigonometry definite-integrals elliptic-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Accordingly, this cannot be solved in terms of elementary functions. In other words, there is no closed form solution and you must resort to numerical integration over a given interval.
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 20:08








  • 3




    $begingroup$
    You have an elliptic integral. Furthermore, at $x = frac {pi}{4}$ the integrand is complex.
    $endgroup$
    – Doug M
    Dec 13 '18 at 20:22












  • $begingroup$
    u = 2.sin x is a nice substitution but results in complex numbers.
    $endgroup$
    – William Elliot
    Dec 13 '18 at 20:30










  • $begingroup$
    You're taking the square root of negative numbers in the integral. Is that what you want?
    $endgroup$
    – zhw.
    Dec 13 '18 at 21:44














4












4








4





$begingroup$


How can we evaluate this integral?
$$int_limits{0}^{pi/4}sqrt{1-16sin^2(x)}mathop{}!mathrm dx$$
I tried a substitution
$$u=4sin x,quad mathrm dx=frac{mathrm du}{sqrt{16-u^2}}$$
hence the integration will be



$$int_limits{u=0}^{u=2sqrt{2}}frac{sqrt{1-u^2}}{sqrt{16-u^2}}mathop{}!mathrm du$$
But I could not complete the solution using this substitution.






calculus integration trigonometry definite-integrals elliptic-integrals







share|cite|improve this question











$endgroup$




How can we evaluate this integral?
$$int_limits{0}^{pi/4}sqrt{1-16sin^2(x)}mathop{}!mathrm dx$$
I tried a substitution
$$u=4sin x,quad mathrm dx=frac{mathrm du}{sqrt{16-u^2}}$$
hence the integration will be



$$int_limits{u=0}^{u=2sqrt{2}}frac{sqrt{1-u^2}}{sqrt{16-u^2}}mathop{}!mathrm du$$
But I could not complete the solution using this substitution.






calculus integration trigonometry definite-integrals elliptic-integrals




calculus integration trigonometry definite-integrals elliptic-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 22:05







user614671

















asked Dec 13 '18 at 20:06









MCSMCS

316211




316211








  • 2




    $begingroup$
    Accordingly, this cannot be solved in terms of elementary functions. In other words, there is no closed form solution and you must resort to numerical integration over a given interval.
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 20:08








  • 3




    $begingroup$
    You have an elliptic integral. Furthermore, at $x = frac {pi}{4}$ the integrand is complex.
    $endgroup$
    – Doug M
    Dec 13 '18 at 20:22












  • $begingroup$
    u = 2.sin x is a nice substitution but results in complex numbers.
    $endgroup$
    – William Elliot
    Dec 13 '18 at 20:30










  • $begingroup$
    You're taking the square root of negative numbers in the integral. Is that what you want?
    $endgroup$
    – zhw.
    Dec 13 '18 at 21:44














  • 2




    $begingroup$
    Accordingly, this cannot be solved in terms of elementary functions. In other words, there is no closed form solution and you must resort to numerical integration over a given interval.
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 20:08








  • 3




    $begingroup$
    You have an elliptic integral. Furthermore, at $x = frac {pi}{4}$ the integrand is complex.
    $endgroup$
    – Doug M
    Dec 13 '18 at 20:22












  • $begingroup$
    u = 2.sin x is a nice substitution but results in complex numbers.
    $endgroup$
    – William Elliot
    Dec 13 '18 at 20:30










  • $begingroup$
    You're taking the square root of negative numbers in the integral. Is that what you want?
    $endgroup$
    – zhw.
    Dec 13 '18 at 21:44








2




2




$begingroup$
Accordingly, this cannot be solved in terms of elementary functions. In other words, there is no closed form solution and you must resort to numerical integration over a given interval.
$endgroup$
– Decaf-Math
Dec 13 '18 at 20:08






$begingroup$
Accordingly, this cannot be solved in terms of elementary functions. In other words, there is no closed form solution and you must resort to numerical integration over a given interval.
$endgroup$
– Decaf-Math
Dec 13 '18 at 20:08






3




3




$begingroup$
You have an elliptic integral. Furthermore, at $x = frac {pi}{4}$ the integrand is complex.
$endgroup$
– Doug M
Dec 13 '18 at 20:22






$begingroup$
You have an elliptic integral. Furthermore, at $x = frac {pi}{4}$ the integrand is complex.
$endgroup$
– Doug M
Dec 13 '18 at 20:22














$begingroup$
u = 2.sin x is a nice substitution but results in complex numbers.
$endgroup$
– William Elliot
Dec 13 '18 at 20:30




$begingroup$
u = 2.sin x is a nice substitution but results in complex numbers.
$endgroup$
– William Elliot
Dec 13 '18 at 20:30












$begingroup$
You're taking the square root of negative numbers in the integral. Is that what you want?
$endgroup$
– zhw.
Dec 13 '18 at 21:44




$begingroup$
You're taking the square root of negative numbers in the integral. Is that what you want?
$endgroup$
– zhw.
Dec 13 '18 at 21:44










1 Answer
1






active

oldest

votes


















5












$begingroup$

begin{align}
int_{0}^{pi/4},sqrt{,{1 - 16sin^{2}left(, x,right)},},,mathrm{d}x & =
int_{0}^{pi/4},sqrt{,{1 - 4^{2}sin^{2}left(, x,right)},},,mathrm{d}x
\[5mm] & =
bbox[10px,#ffd,border:1px groove navy]{mathrm{E}left(,{{pi over 4},4},right)}
end{align}




$displaystylemathrm{E}$ is a
Legendre Integral.







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    @Clayton I don't think so. Maybe, that's ONE of the reasons people define a new "special function".
    $endgroup$
    – Felix Marin
    Dec 13 '18 at 20:46










  • $begingroup$
    Thanks, Felix!${}$
    $endgroup$
    – Clayton
    Dec 13 '18 at 20:51











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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5












$begingroup$

begin{align}
int_{0}^{pi/4},sqrt{,{1 - 16sin^{2}left(, x,right)},},,mathrm{d}x & =
int_{0}^{pi/4},sqrt{,{1 - 4^{2}sin^{2}left(, x,right)},},,mathrm{d}x
\[5mm] & =
bbox[10px,#ffd,border:1px groove navy]{mathrm{E}left(,{{pi over 4},4},right)}
end{align}




$displaystylemathrm{E}$ is a
Legendre Integral.







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    @Clayton I don't think so. Maybe, that's ONE of the reasons people define a new "special function".
    $endgroup$
    – Felix Marin
    Dec 13 '18 at 20:46










  • $begingroup$
    Thanks, Felix!${}$
    $endgroup$
    – Clayton
    Dec 13 '18 at 20:51
















5












$begingroup$

begin{align}
int_{0}^{pi/4},sqrt{,{1 - 16sin^{2}left(, x,right)},},,mathrm{d}x & =
int_{0}^{pi/4},sqrt{,{1 - 4^{2}sin^{2}left(, x,right)},},,mathrm{d}x
\[5mm] & =
bbox[10px,#ffd,border:1px groove navy]{mathrm{E}left(,{{pi over 4},4},right)}
end{align}




$displaystylemathrm{E}$ is a
Legendre Integral.







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    @Clayton I don't think so. Maybe, that's ONE of the reasons people define a new "special function".
    $endgroup$
    – Felix Marin
    Dec 13 '18 at 20:46










  • $begingroup$
    Thanks, Felix!${}$
    $endgroup$
    – Clayton
    Dec 13 '18 at 20:51














5












5








5





$begingroup$

begin{align}
int_{0}^{pi/4},sqrt{,{1 - 16sin^{2}left(, x,right)},},,mathrm{d}x & =
int_{0}^{pi/4},sqrt{,{1 - 4^{2}sin^{2}left(, x,right)},},,mathrm{d}x
\[5mm] & =
bbox[10px,#ffd,border:1px groove navy]{mathrm{E}left(,{{pi over 4},4},right)}
end{align}




$displaystylemathrm{E}$ is a
Legendre Integral.







share|cite|improve this answer









$endgroup$



begin{align}
int_{0}^{pi/4},sqrt{,{1 - 16sin^{2}left(, x,right)},},,mathrm{d}x & =
int_{0}^{pi/4},sqrt{,{1 - 4^{2}sin^{2}left(, x,right)},},,mathrm{d}x
\[5mm] & =
bbox[10px,#ffd,border:1px groove navy]{mathrm{E}left(,{{pi over 4},4},right)}
end{align}




$displaystylemathrm{E}$ is a
Legendre Integral.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 20:38









Felix MarinFelix Marin

68.2k7109143




68.2k7109143








  • 1




    $begingroup$
    @Clayton I don't think so. Maybe, that's ONE of the reasons people define a new "special function".
    $endgroup$
    – Felix Marin
    Dec 13 '18 at 20:46










  • $begingroup$
    Thanks, Felix!${}$
    $endgroup$
    – Clayton
    Dec 13 '18 at 20:51














  • 1




    $begingroup$
    @Clayton I don't think so. Maybe, that's ONE of the reasons people define a new "special function".
    $endgroup$
    – Felix Marin
    Dec 13 '18 at 20:46










  • $begingroup$
    Thanks, Felix!${}$
    $endgroup$
    – Clayton
    Dec 13 '18 at 20:51








1




1




$begingroup$
@Clayton I don't think so. Maybe, that's ONE of the reasons people define a new "special function".
$endgroup$
– Felix Marin
Dec 13 '18 at 20:46




$begingroup$
@Clayton I don't think so. Maybe, that's ONE of the reasons people define a new "special function".
$endgroup$
– Felix Marin
Dec 13 '18 at 20:46












$begingroup$
Thanks, Felix!${}$
$endgroup$
– Clayton
Dec 13 '18 at 20:51




$begingroup$
Thanks, Felix!${}$
$endgroup$
– Clayton
Dec 13 '18 at 20:51


















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