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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038518%2fevaluating-the-integral-int-0-pi-4-sqrt1-16-sin2x-mathop-mathrm-dx%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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
















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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038518%2fevaluating-the-integral-int-0-pi-4-sqrt1-16-sin2x-mathop-mathrm-dx%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Le Mesnil-Réaume

Image
Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more