what is $int_0^1 sqrt{1+2sin^2(t)+cos^2(t)}dt$?












3












$begingroup$


The question is : Is $sqrt 3$ the length of $$Gamma ={gamma (t)=(t,sin(t),sqrt 2cos(t))mid tin [0,1]} ?$$
So it is $$int_0^1|gamma '(t)|dt=int_0^1 sqrt{1+2sin^2(t)+cos^2(t)}dt$$



I tries to do many substitution as $cos^2(t)=frac{1+cos(2t)}{2}$, $sin^2(t)=frac{1-cos(2t)}{2}$, or $1=cos^2(t)+sin^2(t)$ but I can't conclude. I'm sure there is a trick but I don't see it, could someone help ?






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your integral leads to an Elliptic one: $$sqrt {3}{it EllipticE} left( 1/3,sqrt {3} right) -sqrt {3}{it EllipticE} left( cos left( 1 right) ,1/3,sqrt {3} right) $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 7 at 19:14










  • $begingroup$
    This question was ask to a second year of bachelor, so is there an other way to answer ?
    $endgroup$
    – NewMath
    Jan 7 at 19:19










  • $begingroup$
    I don't know, i have only cosidered your integral
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 7 at 19:20
















3












$begingroup$


The question is : Is $sqrt 3$ the length of $$Gamma ={gamma (t)=(t,sin(t),sqrt 2cos(t))mid tin [0,1]} ?$$
So it is $$int_0^1|gamma '(t)|dt=int_0^1 sqrt{1+2sin^2(t)+cos^2(t)}dt$$



I tries to do many substitution as $cos^2(t)=frac{1+cos(2t)}{2}$, $sin^2(t)=frac{1-cos(2t)}{2}$, or $1=cos^2(t)+sin^2(t)$ but I can't conclude. I'm sure there is a trick but I don't see it, could someone help ?






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your integral leads to an Elliptic one: $$sqrt {3}{it EllipticE} left( 1/3,sqrt {3} right) -sqrt {3}{it EllipticE} left( cos left( 1 right) ,1/3,sqrt {3} right) $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 7 at 19:14










  • $begingroup$
    This question was ask to a second year of bachelor, so is there an other way to answer ?
    $endgroup$
    – NewMath
    Jan 7 at 19:19










  • $begingroup$
    I don't know, i have only cosidered your integral
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 7 at 19:20














3












3








3





$begingroup$


The question is : Is $sqrt 3$ the length of $$Gamma ={gamma (t)=(t,sin(t),sqrt 2cos(t))mid tin [0,1]} ?$$
So it is $$int_0^1|gamma '(t)|dt=int_0^1 sqrt{1+2sin^2(t)+cos^2(t)}dt$$



I tries to do many substitution as $cos^2(t)=frac{1+cos(2t)}{2}$, $sin^2(t)=frac{1-cos(2t)}{2}$, or $1=cos^2(t)+sin^2(t)$ but I can't conclude. I'm sure there is a trick but I don't see it, could someone help ?






integration







share|cite|improve this question











$endgroup$




The question is : Is $sqrt 3$ the length of $$Gamma ={gamma (t)=(t,sin(t),sqrt 2cos(t))mid tin [0,1]} ?$$
So it is $$int_0^1|gamma '(t)|dt=int_0^1 sqrt{1+2sin^2(t)+cos^2(t)}dt$$



I tries to do many substitution as $cos^2(t)=frac{1+cos(2t)}{2}$, $sin^2(t)=frac{1-cos(2t)}{2}$, or $1=cos^2(t)+sin^2(t)$ but I can't conclude. I'm sure there is a trick but I don't see it, could someone help ?






integration




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 19:09









clathratus

3,541332




3,541332










asked Jan 7 at 19:08









NewMathNewMath

4029




4029












  • $begingroup$
    Your integral leads to an Elliptic one: $$sqrt {3}{it EllipticE} left( 1/3,sqrt {3} right) -sqrt {3}{it EllipticE} left( cos left( 1 right) ,1/3,sqrt {3} right) $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 7 at 19:14










  • $begingroup$
    This question was ask to a second year of bachelor, so is there an other way to answer ?
    $endgroup$
    – NewMath
    Jan 7 at 19:19










  • $begingroup$
    I don't know, i have only cosidered your integral
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 7 at 19:20


















  • $begingroup$
    Your integral leads to an Elliptic one: $$sqrt {3}{it EllipticE} left( 1/3,sqrt {3} right) -sqrt {3}{it EllipticE} left( cos left( 1 right) ,1/3,sqrt {3} right) $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 7 at 19:14










  • $begingroup$
    This question was ask to a second year of bachelor, so is there an other way to answer ?
    $endgroup$
    – NewMath
    Jan 7 at 19:19










  • $begingroup$
    I don't know, i have only cosidered your integral
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 7 at 19:20
















$begingroup$
Your integral leads to an Elliptic one: $$sqrt {3}{it EllipticE} left( 1/3,sqrt {3} right) -sqrt {3}{it EllipticE} left( cos left( 1 right) ,1/3,sqrt {3} right) $$
$endgroup$
– Dr. Sonnhard Graubner
Jan 7 at 19:14




$begingroup$
Your integral leads to an Elliptic one: $$sqrt {3}{it EllipticE} left( 1/3,sqrt {3} right) -sqrt {3}{it EllipticE} left( cos left( 1 right) ,1/3,sqrt {3} right) $$
$endgroup$
– Dr. Sonnhard Graubner
Jan 7 at 19:14












$begingroup$
This question was ask to a second year of bachelor, so is there an other way to answer ?
$endgroup$
– NewMath
Jan 7 at 19:19




$begingroup$
This question was ask to a second year of bachelor, so is there an other way to answer ?
$endgroup$
– NewMath
Jan 7 at 19:19












$begingroup$
I don't know, i have only cosidered your integral
$endgroup$
– Dr. Sonnhard Graubner
Jan 7 at 19:20




$begingroup$
I don't know, i have only cosidered your integral
$endgroup$
– Dr. Sonnhard Graubner
Jan 7 at 19:20










2 Answers
2






active

oldest

votes


















9












$begingroup$

You are misreading the question that was posed to you. It asks if $sqrt{3}$ can be the value of said integral. It is possible to avoid straight-forward integration by exploring the possibility that it may not. To this end you may be able to find an (easy) inequality to show this.



Notice that $2+sin^2(t) < 3$ for every $tin[0;1]$. Thus:
$$||Gamma||=intlimits_0^1 sqrt{2+sin^2(t)} mathrm{d}t < intlimits_0^1sqrt{3} mathrm{d}t=sqrt{3},$$ which answers your question.






share|cite|improve this answer








New contributor




Snake707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    Very nice ! Good job :)
    $endgroup$
    – NewMath
    Jan 7 at 20:50












  • $begingroup$
    Nice and simple and +1 ! Looking at some comments, other people did misread the question.
    $endgroup$
    – Claude Leibovici
    Jan 8 at 7:01



















3












$begingroup$

$$1=cos^2x+sin^2x$$
$$2=1+cos^2x+sin^2x$$
$$2+sin^2x=1+cos^2x+2sin^2x$$
so
$$||Gamma||=int_0^1sqrt{1+cos^2x+2sin^2x},mathrm dx=int_0^1sqrt{2+sin^2x},mathrm dx$$
Then recall the definition of the incomplete elliptic integral of the second kind
$$mathrm E(phi,k)=int_0^phisqrt{1-ksin^2x},mathrm dx$$
So we have that
$$||Gamma||=sqrt{2},mathrm Ebigg(1,-frac12bigg)approx 1.064728654529...$$
Which might (but probably not) have a closed form






share|cite|improve this answer









$endgroup$













    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%2f3065377%2fwhat-is-int-01-sqrt12-sin2t-cos2tdt%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    You are misreading the question that was posed to you. It asks if $sqrt{3}$ can be the value of said integral. It is possible to avoid straight-forward integration by exploring the possibility that it may not. To this end you may be able to find an (easy) inequality to show this.



    Notice that $2+sin^2(t) < 3$ for every $tin[0;1]$. Thus:
    $$||Gamma||=intlimits_0^1 sqrt{2+sin^2(t)} mathrm{d}t < intlimits_0^1sqrt{3} mathrm{d}t=sqrt{3},$$ which answers your question.






    share|cite|improve this answer








    New contributor




    Snake707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$













    • $begingroup$
      Very nice ! Good job :)
      $endgroup$
      – NewMath
      Jan 7 at 20:50












    • $begingroup$
      Nice and simple and +1 ! Looking at some comments, other people did misread the question.
      $endgroup$
      – Claude Leibovici
      Jan 8 at 7:01
















    9












    $begingroup$

    You are misreading the question that was posed to you. It asks if $sqrt{3}$ can be the value of said integral. It is possible to avoid straight-forward integration by exploring the possibility that it may not. To this end you may be able to find an (easy) inequality to show this.



    Notice that $2+sin^2(t) < 3$ for every $tin[0;1]$. Thus:
    $$||Gamma||=intlimits_0^1 sqrt{2+sin^2(t)} mathrm{d}t < intlimits_0^1sqrt{3} mathrm{d}t=sqrt{3},$$ which answers your question.






    share|cite|improve this answer








    New contributor




    Snake707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$













    • $begingroup$
      Very nice ! Good job :)
      $endgroup$
      – NewMath
      Jan 7 at 20:50












    • $begingroup$
      Nice and simple and +1 ! Looking at some comments, other people did misread the question.
      $endgroup$
      – Claude Leibovici
      Jan 8 at 7:01














    9












    9








    9





    $begingroup$

    You are misreading the question that was posed to you. It asks if $sqrt{3}$ can be the value of said integral. It is possible to avoid straight-forward integration by exploring the possibility that it may not. To this end you may be able to find an (easy) inequality to show this.



    Notice that $2+sin^2(t) < 3$ for every $tin[0;1]$. Thus:
    $$||Gamma||=intlimits_0^1 sqrt{2+sin^2(t)} mathrm{d}t < intlimits_0^1sqrt{3} mathrm{d}t=sqrt{3},$$ which answers your question.






    share|cite|improve this answer








    New contributor




    Snake707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$



    You are misreading the question that was posed to you. It asks if $sqrt{3}$ can be the value of said integral. It is possible to avoid straight-forward integration by exploring the possibility that it may not. To this end you may be able to find an (easy) inequality to show this.



    Notice that $2+sin^2(t) < 3$ for every $tin[0;1]$. Thus:
    $$||Gamma||=intlimits_0^1 sqrt{2+sin^2(t)} mathrm{d}t < intlimits_0^1sqrt{3} mathrm{d}t=sqrt{3},$$ which answers your question.







    share|cite|improve this answer








    New contributor




    Snake707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    share|cite|improve this answer



    share|cite|improve this answer






    New contributor




    Snake707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    answered Jan 7 at 19:44









    Snake707Snake707

    1865




    1865




    New contributor




    Snake707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





    New contributor





    Snake707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    Snake707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.












    • $begingroup$
      Very nice ! Good job :)
      $endgroup$
      – NewMath
      Jan 7 at 20:50












    • $begingroup$
      Nice and simple and +1 ! Looking at some comments, other people did misread the question.
      $endgroup$
      – Claude Leibovici
      Jan 8 at 7:01


















    • $begingroup$
      Very nice ! Good job :)
      $endgroup$
      – NewMath
      Jan 7 at 20:50












    • $begingroup$
      Nice and simple and +1 ! Looking at some comments, other people did misread the question.
      $endgroup$
      – Claude Leibovici
      Jan 8 at 7:01
















    $begingroup$
    Very nice ! Good job :)
    $endgroup$
    – NewMath
    Jan 7 at 20:50






    $begingroup$
    Very nice ! Good job :)
    $endgroup$
    – NewMath
    Jan 7 at 20:50














    $begingroup$
    Nice and simple and +1 ! Looking at some comments, other people did misread the question.
    $endgroup$
    – Claude Leibovici
    Jan 8 at 7:01




    $begingroup$
    Nice and simple and +1 ! Looking at some comments, other people did misread the question.
    $endgroup$
    – Claude Leibovici
    Jan 8 at 7:01











    3












    $begingroup$

    $$1=cos^2x+sin^2x$$
    $$2=1+cos^2x+sin^2x$$
    $$2+sin^2x=1+cos^2x+2sin^2x$$
    so
    $$||Gamma||=int_0^1sqrt{1+cos^2x+2sin^2x},mathrm dx=int_0^1sqrt{2+sin^2x},mathrm dx$$
    Then recall the definition of the incomplete elliptic integral of the second kind
    $$mathrm E(phi,k)=int_0^phisqrt{1-ksin^2x},mathrm dx$$
    So we have that
    $$||Gamma||=sqrt{2},mathrm Ebigg(1,-frac12bigg)approx 1.064728654529...$$
    Which might (but probably not) have a closed form






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      $$1=cos^2x+sin^2x$$
      $$2=1+cos^2x+sin^2x$$
      $$2+sin^2x=1+cos^2x+2sin^2x$$
      so
      $$||Gamma||=int_0^1sqrt{1+cos^2x+2sin^2x},mathrm dx=int_0^1sqrt{2+sin^2x},mathrm dx$$
      Then recall the definition of the incomplete elliptic integral of the second kind
      $$mathrm E(phi,k)=int_0^phisqrt{1-ksin^2x},mathrm dx$$
      So we have that
      $$||Gamma||=sqrt{2},mathrm Ebigg(1,-frac12bigg)approx 1.064728654529...$$
      Which might (but probably not) have a closed form






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        $$1=cos^2x+sin^2x$$
        $$2=1+cos^2x+sin^2x$$
        $$2+sin^2x=1+cos^2x+2sin^2x$$
        so
        $$||Gamma||=int_0^1sqrt{1+cos^2x+2sin^2x},mathrm dx=int_0^1sqrt{2+sin^2x},mathrm dx$$
        Then recall the definition of the incomplete elliptic integral of the second kind
        $$mathrm E(phi,k)=int_0^phisqrt{1-ksin^2x},mathrm dx$$
        So we have that
        $$||Gamma||=sqrt{2},mathrm Ebigg(1,-frac12bigg)approx 1.064728654529...$$
        Which might (but probably not) have a closed form






        share|cite|improve this answer









        $endgroup$



        $$1=cos^2x+sin^2x$$
        $$2=1+cos^2x+sin^2x$$
        $$2+sin^2x=1+cos^2x+2sin^2x$$
        so
        $$||Gamma||=int_0^1sqrt{1+cos^2x+2sin^2x},mathrm dx=int_0^1sqrt{2+sin^2x},mathrm dx$$
        Then recall the definition of the incomplete elliptic integral of the second kind
        $$mathrm E(phi,k)=int_0^phisqrt{1-ksin^2x},mathrm dx$$
        So we have that
        $$||Gamma||=sqrt{2},mathrm Ebigg(1,-frac12bigg)approx 1.064728654529...$$
        Which might (but probably not) have a closed form







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 19:24









        clathratusclathratus

        3,541332




        3,541332






























            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%2f3065377%2fwhat-is-int-01-sqrt12-sin2t-cos2tdt%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

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
            Read more

            Understanding the size os this class of aleatory events

            Image
            0 1 $begingroup$ Hi I am having problems to understand this: I have: $Omega = [0,1]$ , $Lambda =$ all subsets with defined length Let $A_{0}$ = {A $subset$ [0,1]: A is finite union of intervals} 1º: My doubt is that $A_{0}$ is a algebra but its not a sigma algebra. Why? Also, A = $(0,1/2) cup (1/2,3/4) cup (3/4,7/8) cup ... cup(1-1/2^{n},1-1/2^{n+1})...$ 2º: So the class of aleatory events would be bigger than $A_{0}$ . Why? How the set A is greater than $A_{0}$ ? measure-theory share | cite | improve this question asked Jan 8 at 19:17 Diogo Bastos Diogo Bastos 116 6 ...
            Read more

            Partial Derivative Guidance.

            Image
            0 $begingroup$ I was given this tricky problem in which I'm confuse on how to solve it and I would like some help: Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$ . Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives. My attempt (which is wrong): Using chain rule: $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$ $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$ So, $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$ $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$ Thanks, calculus multivariable-calculus share | cite | improve this question edited Jan 19 at 19:38...
            Read more