Not able to integrate











up vote
3
down vote

favorite
1












$displaystyleint_{0}^{pi/2}
frac{sinleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^{2}} ,mathrm{d}x$



i used the property to change reach
$displaystyle 2I =
int_0^{pi/2}frac{sinleft(xright) + cosleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^2} ,mathrm{d}x$
now writing $displaystylesinleft(2xright) =
1 - left[sinleft(xright) - cosleft(xright)right]^{, 2}$, and substituting
$displaystylesinleft(xright) - cosleft(xright) = t$,



how to do further ?.






definite-integrals







share|cite|improve this question
























  • why should we believe it has non closed form result!
    – maveric
    Nov 27 at 18:38










  • There is a probability of 1 that it has no closed form result.
    – Chickenmancer
    Nov 27 at 18:39










  • woow.probability of 1!!
    – maveric
    Nov 27 at 18:47















up vote
3
down vote

favorite
1












$displaystyleint_{0}^{pi/2}
frac{sinleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^{2}} ,mathrm{d}x$



i used the property to change reach
$displaystyle 2I =
int_0^{pi/2}frac{sinleft(xright) + cosleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^2} ,mathrm{d}x$
now writing $displaystylesinleft(2xright) =
1 - left[sinleft(xright) - cosleft(xright)right]^{, 2}$, and substituting
$displaystylesinleft(xright) - cosleft(xright) = t$,



how to do further ?.






definite-integrals







share|cite|improve this question
























  • why should we believe it has non closed form result!
    – maveric
    Nov 27 at 18:38










  • There is a probability of 1 that it has no closed form result.
    – Chickenmancer
    Nov 27 at 18:39










  • woow.probability of 1!!
    – maveric
    Nov 27 at 18:47













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





$displaystyleint_{0}^{pi/2}
frac{sinleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^{2}} ,mathrm{d}x$



i used the property to change reach
$displaystyle 2I =
int_0^{pi/2}frac{sinleft(xright) + cosleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^2} ,mathrm{d}x$
now writing $displaystylesinleft(2xright) =
1 - left[sinleft(xright) - cosleft(xright)right]^{, 2}$, and substituting
$displaystylesinleft(xright) - cosleft(xright) = t$,



how to do further ?.






definite-integrals







share|cite|improve this question















$displaystyleint_{0}^{pi/2}
frac{sinleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^{2}} ,mathrm{d}x$



i used the property to change reach
$displaystyle 2I =
int_0^{pi/2}frac{sinleft(xright) + cosleft(xright)}
{left[1 + ,sqrt{,sinleft(2xright),},right]^2} ,mathrm{d}x$
now writing $displaystylesinleft(2xright) =
1 - left[sinleft(xright) - cosleft(xright)right]^{, 2}$, and substituting
$displaystylesinleft(xright) - cosleft(xright) = t$,



how to do further ?.






definite-integrals




definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 at 20:33









Felix Marin

66.8k7107139




66.8k7107139










asked Nov 27 at 18:09









maveric

63111




63111












  • why should we believe it has non closed form result!
    – maveric
    Nov 27 at 18:38










  • There is a probability of 1 that it has no closed form result.
    – Chickenmancer
    Nov 27 at 18:39










  • woow.probability of 1!!
    – maveric
    Nov 27 at 18:47


















  • why should we believe it has non closed form result!
    – maveric
    Nov 27 at 18:38










  • There is a probability of 1 that it has no closed form result.
    – Chickenmancer
    Nov 27 at 18:39










  • woow.probability of 1!!
    – maveric
    Nov 27 at 18:47
















why should we believe it has non closed form result!
– maveric
Nov 27 at 18:38




why should we believe it has non closed form result!
– maveric
Nov 27 at 18:38












There is a probability of 1 that it has no closed form result.
– Chickenmancer
Nov 27 at 18:39




There is a probability of 1 that it has no closed form result.
– Chickenmancer
Nov 27 at 18:39












woow.probability of 1!!
– maveric
Nov 27 at 18:47




woow.probability of 1!!
– maveric
Nov 27 at 18:47










1 Answer
1






active

oldest

votes

















up vote
6
down vote













As in the OP, the substitution $y=pi/2-x$ gives
$$
begin{aligned}
J&:=
int_0^{pi/2} frac {sin x}{(1+ sqrt{sin 2x})^2} ;dx
\
&=int_0^{pi/2} frac {cos x}{(1+ sqrt{sin 2x})^2} ;dx
text{ ... and thus}
\
&=
frac 12
int_0^{pi/2} frac {sin x+cos x}{(1+ sqrt{sin 2x})^2} ;dxdots
\
&qquadqquadtext{Now formally set $t=sin x-cos x$,}
\
&qquadqquadtext{so $dt=cos x+sin x$,
$t^2=1-2sin xcos x=1-sin 2x$...}
\
&=
frac 12
int_{-1}^{1}
frac {dt}{(1+ sqrt{1-t^2})^2}
=
int_0^1
frac {dt}{(1+ sqrt{1-t^2})^2}
\
&qquadqquadtext{Now use $t=sin u$}
\
&=
int_0^{pi/2}
frac {cos u; du}{(1+ cos u)^2}
\
&qquadqquadtext{Now use $v=tan(u/2)$}
\
&=
int_0^1
frac {frac{1-v^2}{1+v^2}cdotfrac 2{1+v^2}; dv}
{left(frac 2{1+v^2}right)^2}
=
frac 12
int_0^1(1-v^2); dv
\
&=frac 12left[v-frac 13 v^3right]_0^1
=
frac 12cdot frac 23 =frac 13 .
end{aligned}
$$
Computer check, pari/gp:



? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )

%1 = 0.33333333333333333333333333333333333333





share|cite|improve this answer





















    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',
    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%2f3016115%2fnot-able-to-integrate%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








    up vote
    6
    down vote













    As in the OP, the substitution $y=pi/2-x$ gives
    $$
    begin{aligned}
    J&:=
    int_0^{pi/2} frac {sin x}{(1+ sqrt{sin 2x})^2} ;dx
    \
    &=int_0^{pi/2} frac {cos x}{(1+ sqrt{sin 2x})^2} ;dx
    text{ ... and thus}
    \
    &=
    frac 12
    int_0^{pi/2} frac {sin x+cos x}{(1+ sqrt{sin 2x})^2} ;dxdots
    \
    &qquadqquadtext{Now formally set $t=sin x-cos x$,}
    \
    &qquadqquadtext{so $dt=cos x+sin x$,
    $t^2=1-2sin xcos x=1-sin 2x$...}
    \
    &=
    frac 12
    int_{-1}^{1}
    frac {dt}{(1+ sqrt{1-t^2})^2}
    =
    int_0^1
    frac {dt}{(1+ sqrt{1-t^2})^2}
    \
    &qquadqquadtext{Now use $t=sin u$}
    \
    &=
    int_0^{pi/2}
    frac {cos u; du}{(1+ cos u)^2}
    \
    &qquadqquadtext{Now use $v=tan(u/2)$}
    \
    &=
    int_0^1
    frac {frac{1-v^2}{1+v^2}cdotfrac 2{1+v^2}; dv}
    {left(frac 2{1+v^2}right)^2}
    =
    frac 12
    int_0^1(1-v^2); dv
    \
    &=frac 12left[v-frac 13 v^3right]_0^1
    =
    frac 12cdot frac 23 =frac 13 .
    end{aligned}
    $$
    Computer check, pari/gp:



    ? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )
    
%1 = 0.33333333333333333333333333333333333333





    share|cite|improve this answer

























      up vote
      6
      down vote













      As in the OP, the substitution $y=pi/2-x$ gives
      $$
      begin{aligned}
      J&:=
      int_0^{pi/2} frac {sin x}{(1+ sqrt{sin 2x})^2} ;dx
      \
      &=int_0^{pi/2} frac {cos x}{(1+ sqrt{sin 2x})^2} ;dx
      text{ ... and thus}
      \
      &=
      frac 12
      int_0^{pi/2} frac {sin x+cos x}{(1+ sqrt{sin 2x})^2} ;dxdots
      \
      &qquadqquadtext{Now formally set $t=sin x-cos x$,}
      \
      &qquadqquadtext{so $dt=cos x+sin x$,
      $t^2=1-2sin xcos x=1-sin 2x$...}
      \
      &=
      frac 12
      int_{-1}^{1}
      frac {dt}{(1+ sqrt{1-t^2})^2}
      =
      int_0^1
      frac {dt}{(1+ sqrt{1-t^2})^2}
      \
      &qquadqquadtext{Now use $t=sin u$}
      \
      &=
      int_0^{pi/2}
      frac {cos u; du}{(1+ cos u)^2}
      \
      &qquadqquadtext{Now use $v=tan(u/2)$}
      \
      &=
      int_0^1
      frac {frac{1-v^2}{1+v^2}cdotfrac 2{1+v^2}; dv}
      {left(frac 2{1+v^2}right)^2}
      =
      frac 12
      int_0^1(1-v^2); dv
      \
      &=frac 12left[v-frac 13 v^3right]_0^1
      =
      frac 12cdot frac 23 =frac 13 .
      end{aligned}
      $$
      Computer check, pari/gp:



      ? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )
      
%1 = 0.33333333333333333333333333333333333333





      share|cite|improve this answer























        up vote
        6
        down vote










        up vote
        6
        down vote









        As in the OP, the substitution $y=pi/2-x$ gives
        $$
        begin{aligned}
        J&:=
        int_0^{pi/2} frac {sin x}{(1+ sqrt{sin 2x})^2} ;dx
        \
        &=int_0^{pi/2} frac {cos x}{(1+ sqrt{sin 2x})^2} ;dx
        text{ ... and thus}
        \
        &=
        frac 12
        int_0^{pi/2} frac {sin x+cos x}{(1+ sqrt{sin 2x})^2} ;dxdots
        \
        &qquadqquadtext{Now formally set $t=sin x-cos x$,}
        \
        &qquadqquadtext{so $dt=cos x+sin x$,
        $t^2=1-2sin xcos x=1-sin 2x$...}
        \
        &=
        frac 12
        int_{-1}^{1}
        frac {dt}{(1+ sqrt{1-t^2})^2}
        =
        int_0^1
        frac {dt}{(1+ sqrt{1-t^2})^2}
        \
        &qquadqquadtext{Now use $t=sin u$}
        \
        &=
        int_0^{pi/2}
        frac {cos u; du}{(1+ cos u)^2}
        \
        &qquadqquadtext{Now use $v=tan(u/2)$}
        \
        &=
        int_0^1
        frac {frac{1-v^2}{1+v^2}cdotfrac 2{1+v^2}; dv}
        {left(frac 2{1+v^2}right)^2}
        =
        frac 12
        int_0^1(1-v^2); dv
        \
        &=frac 12left[v-frac 13 v^3right]_0^1
        =
        frac 12cdot frac 23 =frac 13 .
        end{aligned}
        $$
        Computer check, pari/gp:



        ? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )
        
%1 = 0.33333333333333333333333333333333333333





        share|cite|improve this answer












        As in the OP, the substitution $y=pi/2-x$ gives
        $$
        begin{aligned}
        J&:=
        int_0^{pi/2} frac {sin x}{(1+ sqrt{sin 2x})^2} ;dx
        \
        &=int_0^{pi/2} frac {cos x}{(1+ sqrt{sin 2x})^2} ;dx
        text{ ... and thus}
        \
        &=
        frac 12
        int_0^{pi/2} frac {sin x+cos x}{(1+ sqrt{sin 2x})^2} ;dxdots
        \
        &qquadqquadtext{Now formally set $t=sin x-cos x$,}
        \
        &qquadqquadtext{so $dt=cos x+sin x$,
        $t^2=1-2sin xcos x=1-sin 2x$...}
        \
        &=
        frac 12
        int_{-1}^{1}
        frac {dt}{(1+ sqrt{1-t^2})^2}
        =
        int_0^1
        frac {dt}{(1+ sqrt{1-t^2})^2}
        \
        &qquadqquadtext{Now use $t=sin u$}
        \
        &=
        int_0^{pi/2}
        frac {cos u; du}{(1+ cos u)^2}
        \
        &qquadqquadtext{Now use $v=tan(u/2)$}
        \
        &=
        int_0^1
        frac {frac{1-v^2}{1+v^2}cdotfrac 2{1+v^2}; dv}
        {left(frac 2{1+v^2}right)^2}
        =
        frac 12
        int_0^1(1-v^2); dv
        \
        &=frac 12left[v-frac 13 v^3right]_0^1
        =
        frac 12cdot frac 23 =frac 13 .
        end{aligned}
        $$
        Computer check, pari/gp:



        ? intnum( x=0, Pi/2, sin(x) / ( 1+sqrt(sin(2*x)) )^2 )
        
%1 = 0.33333333333333333333333333333333333333






        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 at 18:46









        dan_fulea

        6,2201312




        6,2201312






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3016115%2fnot-able-to-integrate%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

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
            Read more

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
            Read more

            Sphinx de Gizeh

            Image
            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
            Read more