Integral: $intfrac{sin^3theta}{sqrt{costheta-2}},dtheta$












-1














I tried using the substitution of $u=costheta-2$ which gives $dfrac{dtheta}{du}=dfrac{-1}{sintheta}$. Then,



$displaystyleintdfrac{sin^3theta}{sqrt{costheta-2}},dtheta=displaystyleintdfrac{sin^2thetasintheta}{sqrt{u}}left(dfrac{-1}{sintheta}right)du=-displaystyleintdfrac{sin^2theta}{sqrt{u}}du$.



I am stuck here...perhaps using $sin^2theta=1-cos^2theta$ might help, but I don't know how.






integration indefinite-integrals trigonometric-integrals







share|cite|improve this question




















  • 2




    If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
    – Beni Bogosel
    Nov 18 '14 at 11:54










  • Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
    – bibo_extreme
    Nov 18 '14 at 11:57










  • You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
    – Beni Bogosel
    Nov 18 '14 at 11:58










  • Most likely, yes.
    – bibo_extreme
    Nov 18 '14 at 12:00
















-1














I tried using the substitution of $u=costheta-2$ which gives $dfrac{dtheta}{du}=dfrac{-1}{sintheta}$. Then,



$displaystyleintdfrac{sin^3theta}{sqrt{costheta-2}},dtheta=displaystyleintdfrac{sin^2thetasintheta}{sqrt{u}}left(dfrac{-1}{sintheta}right)du=-displaystyleintdfrac{sin^2theta}{sqrt{u}}du$.



I am stuck here...perhaps using $sin^2theta=1-cos^2theta$ might help, but I don't know how.






integration indefinite-integrals trigonometric-integrals







share|cite|improve this question




















  • 2




    If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
    – Beni Bogosel
    Nov 18 '14 at 11:54










  • Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
    – bibo_extreme
    Nov 18 '14 at 11:57










  • You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
    – Beni Bogosel
    Nov 18 '14 at 11:58










  • Most likely, yes.
    – bibo_extreme
    Nov 18 '14 at 12:00














-1












-1








-1







I tried using the substitution of $u=costheta-2$ which gives $dfrac{dtheta}{du}=dfrac{-1}{sintheta}$. Then,



$displaystyleintdfrac{sin^3theta}{sqrt{costheta-2}},dtheta=displaystyleintdfrac{sin^2thetasintheta}{sqrt{u}}left(dfrac{-1}{sintheta}right)du=-displaystyleintdfrac{sin^2theta}{sqrt{u}}du$.



I am stuck here...perhaps using $sin^2theta=1-cos^2theta$ might help, but I don't know how.






integration indefinite-integrals trigonometric-integrals







share|cite|improve this question















I tried using the substitution of $u=costheta-2$ which gives $dfrac{dtheta}{du}=dfrac{-1}{sintheta}$. Then,



$displaystyleintdfrac{sin^3theta}{sqrt{costheta-2}},dtheta=displaystyleintdfrac{sin^2thetasintheta}{sqrt{u}}left(dfrac{-1}{sintheta}right)du=-displaystyleintdfrac{sin^2theta}{sqrt{u}}du$.



I am stuck here...perhaps using $sin^2theta=1-cos^2theta$ might help, but I don't know how.






integration indefinite-integrals trigonometric-integrals




integration indefinite-integrals trigonometric-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 17:34









Martin Sleziak

44.6k7115270




44.6k7115270










asked Nov 18 '14 at 11:50









bibo_extreme

337110




337110








  • 2




    If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
    – Beni Bogosel
    Nov 18 '14 at 11:54










  • Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
    – bibo_extreme
    Nov 18 '14 at 11:57










  • You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
    – Beni Bogosel
    Nov 18 '14 at 11:58










  • Most likely, yes.
    – bibo_extreme
    Nov 18 '14 at 12:00














  • 2




    If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
    – Beni Bogosel
    Nov 18 '14 at 11:54










  • Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
    – bibo_extreme
    Nov 18 '14 at 11:57










  • You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
    – Beni Bogosel
    Nov 18 '14 at 11:58










  • Most likely, yes.
    – bibo_extreme
    Nov 18 '14 at 12:00








2




2




If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
– Beni Bogosel
Nov 18 '14 at 11:54




If you are working with real numbers, then $sqrt{cos theta -2}$ makes no sense.
– Beni Bogosel
Nov 18 '14 at 11:54












Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
– bibo_extreme
Nov 18 '14 at 11:57




Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me.
– bibo_extreme
Nov 18 '14 at 11:57












You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
– Beni Bogosel
Nov 18 '14 at 11:58




You could replace it with $sqrt{2-cos theta}$. Maybe that's what the original question had.
– Beni Bogosel
Nov 18 '14 at 11:58












Most likely, yes.
– bibo_extreme
Nov 18 '14 at 12:00




Most likely, yes.
– bibo_extreme
Nov 18 '14 at 12:00










3 Answers
3






active

oldest

votes


















2














As Beni points out, this only makes sense in the complex setting (and with a choice of branch cut), but you can put the integral in terms of $u$ alone via the Pythagorean Identity:
$$sin^2 theta = 1 - cos^2 theta = 1 - (u + 2)^2 = -(u^2 + 4u + 3).$$






share|cite|improve this answer































    2














    Let $sqrt{costheta-2}=uimpliesdfrac{-sintheta dtheta}{sqrt{costheta-2}}=du$



    and $costheta=u^2+2$



    $$intfrac{sin^3theta}{sqrt{costheta-2}}dtheta=-int[1-(u^2+2)^2]du$$






    share|cite|improve this answer





























      1














      Use $sin{theta}, dtheta = -d(cos{theta})$. Then the integral is



      $$iint d(cos{theta}) frac{sin^2{theta}}{sqrt{2-cos{theta}}}$$



      (Note the $i$ comes from the square root.)



      Let $u=cos{theta}$; then we have



      $$i int du frac{1-u^2}{sqrt{2-u}} = -i int dv frac{1-(2-v)^2}{sqrt{v}} = i int dv , v^{-1/2} (3 - 4 v+v^2)$$



      I think you can take it from here.






      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%2f1027460%2fintegral-int-frac-sin3-theta-sqrt-cos-theta-2-d-theta%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2














        As Beni points out, this only makes sense in the complex setting (and with a choice of branch cut), but you can put the integral in terms of $u$ alone via the Pythagorean Identity:
        $$sin^2 theta = 1 - cos^2 theta = 1 - (u + 2)^2 = -(u^2 + 4u + 3).$$






        share|cite|improve this answer




























          2














          As Beni points out, this only makes sense in the complex setting (and with a choice of branch cut), but you can put the integral in terms of $u$ alone via the Pythagorean Identity:
          $$sin^2 theta = 1 - cos^2 theta = 1 - (u + 2)^2 = -(u^2 + 4u + 3).$$






          share|cite|improve this answer


























            2












            2








            2






            As Beni points out, this only makes sense in the complex setting (and with a choice of branch cut), but you can put the integral in terms of $u$ alone via the Pythagorean Identity:
            $$sin^2 theta = 1 - cos^2 theta = 1 - (u + 2)^2 = -(u^2 + 4u + 3).$$






            share|cite|improve this answer














            As Beni points out, this only makes sense in the complex setting (and with a choice of branch cut), but you can put the integral in terms of $u$ alone via the Pythagorean Identity:
            $$sin^2 theta = 1 - cos^2 theta = 1 - (u + 2)^2 = -(u^2 + 4u + 3).$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 18 '14 at 12:25

























            answered Nov 18 '14 at 11:53









            Travis

            59.4k767145




            59.4k767145























                2














                Let $sqrt{costheta-2}=uimpliesdfrac{-sintheta dtheta}{sqrt{costheta-2}}=du$



                and $costheta=u^2+2$



                $$intfrac{sin^3theta}{sqrt{costheta-2}}dtheta=-int[1-(u^2+2)^2]du$$






                share|cite|improve this answer


























                  2














                  Let $sqrt{costheta-2}=uimpliesdfrac{-sintheta dtheta}{sqrt{costheta-2}}=du$



                  and $costheta=u^2+2$



                  $$intfrac{sin^3theta}{sqrt{costheta-2}}dtheta=-int[1-(u^2+2)^2]du$$






                  share|cite|improve this answer
























                    2












                    2








                    2






                    Let $sqrt{costheta-2}=uimpliesdfrac{-sintheta dtheta}{sqrt{costheta-2}}=du$



                    and $costheta=u^2+2$



                    $$intfrac{sin^3theta}{sqrt{costheta-2}}dtheta=-int[1-(u^2+2)^2]du$$






                    share|cite|improve this answer












                    Let $sqrt{costheta-2}=uimpliesdfrac{-sintheta dtheta}{sqrt{costheta-2}}=du$



                    and $costheta=u^2+2$



                    $$intfrac{sin^3theta}{sqrt{costheta-2}}dtheta=-int[1-(u^2+2)^2]du$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 18 '14 at 11:59









                    lab bhattacharjee

                    222k15155273




                    222k15155273























                        1














                        Use $sin{theta}, dtheta = -d(cos{theta})$. Then the integral is



                        $$iint d(cos{theta}) frac{sin^2{theta}}{sqrt{2-cos{theta}}}$$



                        (Note the $i$ comes from the square root.)



                        Let $u=cos{theta}$; then we have



                        $$i int du frac{1-u^2}{sqrt{2-u}} = -i int dv frac{1-(2-v)^2}{sqrt{v}} = i int dv , v^{-1/2} (3 - 4 v+v^2)$$



                        I think you can take it from here.






                        share|cite|improve this answer


























                          1














                          Use $sin{theta}, dtheta = -d(cos{theta})$. Then the integral is



                          $$iint d(cos{theta}) frac{sin^2{theta}}{sqrt{2-cos{theta}}}$$



                          (Note the $i$ comes from the square root.)



                          Let $u=cos{theta}$; then we have



                          $$i int du frac{1-u^2}{sqrt{2-u}} = -i int dv frac{1-(2-v)^2}{sqrt{v}} = i int dv , v^{-1/2} (3 - 4 v+v^2)$$



                          I think you can take it from here.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            Use $sin{theta}, dtheta = -d(cos{theta})$. Then the integral is



                            $$iint d(cos{theta}) frac{sin^2{theta}}{sqrt{2-cos{theta}}}$$



                            (Note the $i$ comes from the square root.)



                            Let $u=cos{theta}$; then we have



                            $$i int du frac{1-u^2}{sqrt{2-u}} = -i int dv frac{1-(2-v)^2}{sqrt{v}} = i int dv , v^{-1/2} (3 - 4 v+v^2)$$



                            I think you can take it from here.






                            share|cite|improve this answer












                            Use $sin{theta}, dtheta = -d(cos{theta})$. Then the integral is



                            $$iint d(cos{theta}) frac{sin^2{theta}}{sqrt{2-cos{theta}}}$$



                            (Note the $i$ comes from the square root.)



                            Let $u=cos{theta}$; then we have



                            $$i int du frac{1-u^2}{sqrt{2-u}} = -i int dv frac{1-(2-v)^2}{sqrt{v}} = i int dv , v^{-1/2} (3 - 4 v+v^2)$$



                            I think you can take it from here.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 18 '14 at 11:55









                            Ron Gordon

                            122k14154263




                            122k14154263






























                                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%2f1027460%2fintegral-int-frac-sin3-theta-sqrt-cos-theta-2-d-theta%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

                                Verónica Boquete

                                Image
                                Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                                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