$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)},dx$












0














I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$



I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.






complex-analysis contour-integration







share|cite|improve this question




















  • 1




    I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
    – Robert Israel
    2 days ago










  • Oh true, ill give that a try. Thanks
    – Sam
    2 days ago










  • Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
    – Dr. Sonnhard Graubner
    2 days ago
















0














I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$



I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.






complex-analysis contour-integration







share|cite|improve this question




















  • 1




    I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
    – Robert Israel
    2 days ago










  • Oh true, ill give that a try. Thanks
    – Sam
    2 days ago










  • Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
    – Dr. Sonnhard Graubner
    2 days ago














0












0








0


1





I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$



I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.






complex-analysis contour-integration







share|cite|improve this question















I have been trying to solve the integral below using contour integration.
$$int_{0}^{pi}frac{sin^2(x)}{a+cos(x)}dx, quad a>1.$$



I'd appreciate to see how you would solve it, since the fact that it goes to only $pi$ and not to $2pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.






complex-analysis contour-integration




complex-analysis contour-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Bernard

118k639112




118k639112










asked 2 days ago









Sam

372114




372114








  • 1




    I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
    – Robert Israel
    2 days ago










  • Oh true, ill give that a try. Thanks
    – Sam
    2 days ago










  • Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
    – Dr. Sonnhard Graubner
    2 days ago














  • 1




    I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
    – Robert Israel
    2 days ago










  • Oh true, ill give that a try. Thanks
    – Sam
    2 days ago










  • Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
    – Dr. Sonnhard Graubner
    2 days ago








1




1




I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
– Robert Israel
2 days ago




I presume you mean $dx$, not $dt$. Note that the integrand is an even function, thus you can integrate from $-pi$ to $pi$ and divide by $2$.
– Robert Israel
2 days ago












Oh true, ill give that a try. Thanks
– Sam
2 days ago




Oh true, ill give that a try. Thanks
– Sam
2 days ago












Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
– Dr. Sonnhard Graubner
2 days ago




Use this substitution de.wikipedia.org/wiki/Weierstraß-Substitution
– Dr. Sonnhard Graubner
2 days ago










1 Answer
1






active

oldest

votes


















3














The chances of me saying this were high: let us avoid contour integration.

By symmetry
$$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
equals
$$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
or
$$ frac{pi}{a+sqrt{a^2-1}} $$
by partial fraction decomposition.






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',
    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%2f3060749%2fint-0-pi-frac-sin2xa-cosx-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









    3














    The chances of me saying this were high: let us avoid contour integration.

    By symmetry
    $$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
    equals
    $$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
    or
    $$ frac{pi}{a+sqrt{a^2-1}} $$
    by partial fraction decomposition.






    share|cite|improve this answer


























      3














      The chances of me saying this were high: let us avoid contour integration.

      By symmetry
      $$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
      equals
      $$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
      or
      $$ frac{pi}{a+sqrt{a^2-1}} $$
      by partial fraction decomposition.






      share|cite|improve this answer
























        3












        3








        3






        The chances of me saying this were high: let us avoid contour integration.

        By symmetry
        $$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
        equals
        $$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
        or
        $$ frac{pi}{a+sqrt{a^2-1}} $$
        by partial fraction decomposition.






        share|cite|improve this answer












        The chances of me saying this were high: let us avoid contour integration.

        By symmetry
        $$ int_{0}^{pi}frac{sin^2theta}{a+costheta},dtheta = int_{0}^{pi/2}sin^2(theta)left[frac{1}{a+costheta}+frac{1}{a-costheta}right]dtheta$$
        equals
        $$ 2aint_{0}^{pi/2}frac{sin^2theta}{a^2-cos^2theta},dthetastackrel{thetamapstoarctan t}{=}2aint_{0}^{+infty}frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$
        or
        $$ frac{pi}{a+sqrt{a^2-1}} $$
        by partial fraction decomposition.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Jack D'Aurizio

        287k33280657




        287k33280657






























            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%2f3060749%2fint-0-pi-frac-sin2xa-cosx-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

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            Display a custom attribute below product name in the front-end Magento 1.9.3.8

            Image
            0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
            Read more