Integral of $frac{x^2}{sqrt{x^2+5}}$












1












$begingroup$


I need help with this integral:



$$I = int frac{x^2}{sqrt{x^2+5}}, dx$$



I substituted $x = sqrt{5}tan{theta}$, and reached $$I = 5int frac{sin{(theta)}^2}{cos{(theta)}^3},dtheta$$ which I'm unable to solve.
For context this is in an exercise of trig substitution, and I'm not allowed to use the secant reduction formula which WolframAlpha suggests I use.






calculus integration trigonometry







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I need help with this integral:



    $$I = int frac{x^2}{sqrt{x^2+5}}, dx$$



    I substituted $x = sqrt{5}tan{theta}$, and reached $$I = 5int frac{sin{(theta)}^2}{cos{(theta)}^3},dtheta$$ which I'm unable to solve.
    For context this is in an exercise of trig substitution, and I'm not allowed to use the secant reduction formula which WolframAlpha suggests I use.






    calculus integration trigonometry







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I need help with this integral:



      $$I = int frac{x^2}{sqrt{x^2+5}}, dx$$



      I substituted $x = sqrt{5}tan{theta}$, and reached $$I = 5int frac{sin{(theta)}^2}{cos{(theta)}^3},dtheta$$ which I'm unable to solve.
      For context this is in an exercise of trig substitution, and I'm not allowed to use the secant reduction formula which WolframAlpha suggests I use.






      calculus integration trigonometry







      share|cite|improve this question











      $endgroup$




      I need help with this integral:



      $$I = int frac{x^2}{sqrt{x^2+5}}, dx$$



      I substituted $x = sqrt{5}tan{theta}$, and reached $$I = 5int frac{sin{(theta)}^2}{cos{(theta)}^3},dtheta$$ which I'm unable to solve.
      For context this is in an exercise of trig substitution, and I'm not allowed to use the secant reduction formula which WolframAlpha suggests I use.






      calculus integration trigonometry




      calculus integration trigonometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 18 at 13:28









      Robert Z

      97.4k1066137




      97.4k1066137










      asked Jan 18 at 12:55









      Yizhar AmirYizhar Amir

      4216




      4216






















          6 Answers
          6






          active

          oldest

          votes


















          3












          $begingroup$

          Hint: (Use hyperbolic trigonometric substitution to)
          $$
          frac{x^2}{sqrt{x^2+5}}=sqrt{x^2+5}-frac{5}{sqrt{x^2+5}}.
          $$






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            Note that$$frac{sin^2(theta)}{cos^3(theta)}=frac{sin^2(theta)cos(theta)}{bigl(1-sin^2(theta)bigr)^2}.$$So, you can do $sin(theta)=u$ and $cos(theta),mathrm dtheta=mathrm du$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
              $endgroup$
              – GEdgar
              Jan 18 at 13:08



















            2












            $begingroup$

            Hint. Let $t$ such that $x^2+5=(x-t)^2$, that is $x=(t^2-5)/(2t)$. Then new integral is the integral of a rational function (very easy to calculate):
            $$int frac{x^2}{sqrt{x^2+5}}, dx=int left(frac{25}{4t^3}+frac{t}{4}-frac{5}{2t}right), dt.$$
            Finally note that $t=x+sqrt{x^2+5}$.






            share|cite|improve this answer











            $endgroup$





















              1












              $begingroup$

              Hint: Rewrite $intfrac{sin^2{(theta)}}{cos^3{(theta)}} d theta$ as $int tan^2 theta sec theta d theta$. Let $u = sec theta$; then $du = ?$ - this will reduce the integral to something trivial.






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                Hint: Substitute $y=sqrt{x^2+5}$. Then $dy= frac{x}{sqrt{x^2+5}}dx$, so the integral is $int sqrt{y^2-5} dy$.






                share|cite|improve this answer









                $endgroup$





















                  0












                  $begingroup$

                  What about



                  $$1+frac x{sqrt5}=sinh uimplies dx=cosh usqrt 5,duimplies$$



                  $$intfrac{x^2}{sqrt{x^2+5}}==frac1{sqrt5}intfrac{x^2}{sqrt{1+left(frac x{sqrt 5}right)^2}}dx=frac1{sqrt5}intfrac{(sqrt5,sinh u-1)^2}{cosh u}sqrt5,cosh u,du=$$



                  $$intleft(5sinh^2u-10sinh u+1right)du$$



                  and now it is almost an immediate integral.






                  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%2f3078210%2fintegral-of-fracx2-sqrtx25%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown

























                    6 Answers
                    6






                    active

                    oldest

                    votes








                    6 Answers
                    6






                    active

                    oldest

                    votes









                    active

                    oldest

                    votes






                    active

                    oldest

                    votes









                    3












                    $begingroup$

                    Hint: (Use hyperbolic trigonometric substitution to)
                    $$
                    frac{x^2}{sqrt{x^2+5}}=sqrt{x^2+5}-frac{5}{sqrt{x^2+5}}.
                    $$






                    share|cite|improve this answer









                    $endgroup$


















                      3












                      $begingroup$

                      Hint: (Use hyperbolic trigonometric substitution to)
                      $$
                      frac{x^2}{sqrt{x^2+5}}=sqrt{x^2+5}-frac{5}{sqrt{x^2+5}}.
                      $$






                      share|cite|improve this answer









                      $endgroup$
















                        3












                        3








                        3





                        $begingroup$

                        Hint: (Use hyperbolic trigonometric substitution to)
                        $$
                        frac{x^2}{sqrt{x^2+5}}=sqrt{x^2+5}-frac{5}{sqrt{x^2+5}}.
                        $$






                        share|cite|improve this answer









                        $endgroup$



                        Hint: (Use hyperbolic trigonometric substitution to)
                        $$
                        frac{x^2}{sqrt{x^2+5}}=sqrt{x^2+5}-frac{5}{sqrt{x^2+5}}.
                        $$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 18 at 12:59









                        SongSong

                        12.9k631




                        12.9k631























                            3












                            $begingroup$

                            Note that$$frac{sin^2(theta)}{cos^3(theta)}=frac{sin^2(theta)cos(theta)}{bigl(1-sin^2(theta)bigr)^2}.$$So, you can do $sin(theta)=u$ and $cos(theta),mathrm dtheta=mathrm du$.






                            share|cite|improve this answer









                            $endgroup$













                            • $begingroup$
                              This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                              $endgroup$
                              – GEdgar
                              Jan 18 at 13:08
















                            3












                            $begingroup$

                            Note that$$frac{sin^2(theta)}{cos^3(theta)}=frac{sin^2(theta)cos(theta)}{bigl(1-sin^2(theta)bigr)^2}.$$So, you can do $sin(theta)=u$ and $cos(theta),mathrm dtheta=mathrm du$.






                            share|cite|improve this answer









                            $endgroup$













                            • $begingroup$
                              This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                              $endgroup$
                              – GEdgar
                              Jan 18 at 13:08














                            3












                            3








                            3





                            $begingroup$

                            Note that$$frac{sin^2(theta)}{cos^3(theta)}=frac{sin^2(theta)cos(theta)}{bigl(1-sin^2(theta)bigr)^2}.$$So, you can do $sin(theta)=u$ and $cos(theta),mathrm dtheta=mathrm du$.






                            share|cite|improve this answer









                            $endgroup$



                            Note that$$frac{sin^2(theta)}{cos^3(theta)}=frac{sin^2(theta)cos(theta)}{bigl(1-sin^2(theta)bigr)^2}.$$So, you can do $sin(theta)=u$ and $cos(theta),mathrm dtheta=mathrm du$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 18 at 13:03









                            José Carlos SantosJosé Carlos Santos

                            161k22127232




                            161k22127232












                            • $begingroup$
                              This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                              $endgroup$
                              – GEdgar
                              Jan 18 at 13:08


















                            • $begingroup$
                              This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                              $endgroup$
                              – GEdgar
                              Jan 18 at 13:08
















                            $begingroup$
                            This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                            $endgroup$
                            – GEdgar
                            Jan 18 at 13:08




                            $begingroup$
                            This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                            $endgroup$
                            – GEdgar
                            Jan 18 at 13:08











                            2












                            $begingroup$

                            Hint. Let $t$ such that $x^2+5=(x-t)^2$, that is $x=(t^2-5)/(2t)$. Then new integral is the integral of a rational function (very easy to calculate):
                            $$int frac{x^2}{sqrt{x^2+5}}, dx=int left(frac{25}{4t^3}+frac{t}{4}-frac{5}{2t}right), dt.$$
                            Finally note that $t=x+sqrt{x^2+5}$.






                            share|cite|improve this answer











                            $endgroup$


















                              2












                              $begingroup$

                              Hint. Let $t$ such that $x^2+5=(x-t)^2$, that is $x=(t^2-5)/(2t)$. Then new integral is the integral of a rational function (very easy to calculate):
                              $$int frac{x^2}{sqrt{x^2+5}}, dx=int left(frac{25}{4t^3}+frac{t}{4}-frac{5}{2t}right), dt.$$
                              Finally note that $t=x+sqrt{x^2+5}$.






                              share|cite|improve this answer











                              $endgroup$
















                                2












                                2








                                2





                                $begingroup$

                                Hint. Let $t$ such that $x^2+5=(x-t)^2$, that is $x=(t^2-5)/(2t)$. Then new integral is the integral of a rational function (very easy to calculate):
                                $$int frac{x^2}{sqrt{x^2+5}}, dx=int left(frac{25}{4t^3}+frac{t}{4}-frac{5}{2t}right), dt.$$
                                Finally note that $t=x+sqrt{x^2+5}$.






                                share|cite|improve this answer











                                $endgroup$



                                Hint. Let $t$ such that $x^2+5=(x-t)^2$, that is $x=(t^2-5)/(2t)$. Then new integral is the integral of a rational function (very easy to calculate):
                                $$int frac{x^2}{sqrt{x^2+5}}, dx=int left(frac{25}{4t^3}+frac{t}{4}-frac{5}{2t}right), dt.$$
                                Finally note that $t=x+sqrt{x^2+5}$.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Jan 18 at 13:26

























                                answered Jan 18 at 13:08









                                Robert ZRobert Z

                                97.4k1066137




                                97.4k1066137























                                    1












                                    $begingroup$

                                    Hint: Rewrite $intfrac{sin^2{(theta)}}{cos^3{(theta)}} d theta$ as $int tan^2 theta sec theta d theta$. Let $u = sec theta$; then $du = ?$ - this will reduce the integral to something trivial.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      1












                                      $begingroup$

                                      Hint: Rewrite $intfrac{sin^2{(theta)}}{cos^3{(theta)}} d theta$ as $int tan^2 theta sec theta d theta$. Let $u = sec theta$; then $du = ?$ - this will reduce the integral to something trivial.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        1












                                        1








                                        1





                                        $begingroup$

                                        Hint: Rewrite $intfrac{sin^2{(theta)}}{cos^3{(theta)}} d theta$ as $int tan^2 theta sec theta d theta$. Let $u = sec theta$; then $du = ?$ - this will reduce the integral to something trivial.






                                        share|cite|improve this answer









                                        $endgroup$



                                        Hint: Rewrite $intfrac{sin^2{(theta)}}{cos^3{(theta)}} d theta$ as $int tan^2 theta sec theta d theta$. Let $u = sec theta$; then $du = ?$ - this will reduce the integral to something trivial.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Jan 18 at 13:43









                                        bjcolby15bjcolby15

                                        1,36711016




                                        1,36711016























                                            0












                                            $begingroup$

                                            Hint: Substitute $y=sqrt{x^2+5}$. Then $dy= frac{x}{sqrt{x^2+5}}dx$, so the integral is $int sqrt{y^2-5} dy$.






                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$

                                              Hint: Substitute $y=sqrt{x^2+5}$. Then $dy= frac{x}{sqrt{x^2+5}}dx$, so the integral is $int sqrt{y^2-5} dy$.






                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                Hint: Substitute $y=sqrt{x^2+5}$. Then $dy= frac{x}{sqrt{x^2+5}}dx$, so the integral is $int sqrt{y^2-5} dy$.






                                                share|cite|improve this answer









                                                $endgroup$



                                                Hint: Substitute $y=sqrt{x^2+5}$. Then $dy= frac{x}{sqrt{x^2+5}}dx$, so the integral is $int sqrt{y^2-5} dy$.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Jan 18 at 13:01









                                                A. PongráczA. Pongrácz

                                                5,9531929




                                                5,9531929























                                                    0












                                                    $begingroup$

                                                    What about



                                                    $$1+frac x{sqrt5}=sinh uimplies dx=cosh usqrt 5,duimplies$$



                                                    $$intfrac{x^2}{sqrt{x^2+5}}==frac1{sqrt5}intfrac{x^2}{sqrt{1+left(frac x{sqrt 5}right)^2}}dx=frac1{sqrt5}intfrac{(sqrt5,sinh u-1)^2}{cosh u}sqrt5,cosh u,du=$$



                                                    $$intleft(5sinh^2u-10sinh u+1right)du$$



                                                    and now it is almost an immediate integral.






                                                    share|cite|improve this answer









                                                    $endgroup$


















                                                      0












                                                      $begingroup$

                                                      What about



                                                      $$1+frac x{sqrt5}=sinh uimplies dx=cosh usqrt 5,duimplies$$



                                                      $$intfrac{x^2}{sqrt{x^2+5}}==frac1{sqrt5}intfrac{x^2}{sqrt{1+left(frac x{sqrt 5}right)^2}}dx=frac1{sqrt5}intfrac{(sqrt5,sinh u-1)^2}{cosh u}sqrt5,cosh u,du=$$



                                                      $$intleft(5sinh^2u-10sinh u+1right)du$$



                                                      and now it is almost an immediate integral.






                                                      share|cite|improve this answer









                                                      $endgroup$
















                                                        0












                                                        0








                                                        0





                                                        $begingroup$

                                                        What about



                                                        $$1+frac x{sqrt5}=sinh uimplies dx=cosh usqrt 5,duimplies$$



                                                        $$intfrac{x^2}{sqrt{x^2+5}}==frac1{sqrt5}intfrac{x^2}{sqrt{1+left(frac x{sqrt 5}right)^2}}dx=frac1{sqrt5}intfrac{(sqrt5,sinh u-1)^2}{cosh u}sqrt5,cosh u,du=$$



                                                        $$intleft(5sinh^2u-10sinh u+1right)du$$



                                                        and now it is almost an immediate integral.






                                                        share|cite|improve this answer









                                                        $endgroup$



                                                        What about



                                                        $$1+frac x{sqrt5}=sinh uimplies dx=cosh usqrt 5,duimplies$$



                                                        $$intfrac{x^2}{sqrt{x^2+5}}==frac1{sqrt5}intfrac{x^2}{sqrt{1+left(frac x{sqrt 5}right)^2}}dx=frac1{sqrt5}intfrac{(sqrt5,sinh u-1)^2}{cosh u}sqrt5,cosh u,du=$$



                                                        $$intleft(5sinh^2u-10sinh u+1right)du$$



                                                        and now it is almost an immediate integral.







                                                        share|cite|improve this answer












                                                        share|cite|improve this answer



                                                        share|cite|improve this answer










                                                        answered Jan 18 at 13:03









                                                        DonAntonioDonAntonio

                                                        178k1494230




                                                        178k1494230






























                                                            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%2f3078210%2fintegral-of-fracx2-sqrtx25%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

                                                            What does “Dominus providebit” mean?

                                                            Image
                                                            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
                                                            Read more

                                                            Antonio Litta Visconti Arese

                                                            Image
                                                            Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
                                                            Read more