Integral by trigonometric substitution: $ int :frac{2y^3}{sqrt{1-y^2}}mathrm dy $












1












$begingroup$



$$ int :frac{2y^3}{sqrt{1-y^2}}mathrm dy $$




I substitute $$ y = sin t $$ then $$ 2 int :frac{sin^3 t cdot cos t}{cos t};mathrm dt = 2 int sin^2 t;mathrm d cos t $$
What should I do next, and how do I convert back to the original variable $y$?






calculus integration substitution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why not $$sin t=y$$
    $endgroup$
    – lab bhattacharjee
    Dec 16 '18 at 18:41






  • 1




    $begingroup$
    Another way: $$sqrt{1-y^2}=u$$
    $endgroup$
    – lab bhattacharjee
    Dec 16 '18 at 18:41






  • 1




    $begingroup$
    $sin^2t=1-cos^2t$.
    $endgroup$
    – Yves Daoust
    Dec 16 '18 at 20:03
















1












$begingroup$



$$ int :frac{2y^3}{sqrt{1-y^2}}mathrm dy $$




I substitute $$ y = sin t $$ then $$ 2 int :frac{sin^3 t cdot cos t}{cos t};mathrm dt = 2 int sin^2 t;mathrm d cos t $$
What should I do next, and how do I convert back to the original variable $y$?






calculus integration substitution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why not $$sin t=y$$
    $endgroup$
    – lab bhattacharjee
    Dec 16 '18 at 18:41






  • 1




    $begingroup$
    Another way: $$sqrt{1-y^2}=u$$
    $endgroup$
    – lab bhattacharjee
    Dec 16 '18 at 18:41






  • 1




    $begingroup$
    $sin^2t=1-cos^2t$.
    $endgroup$
    – Yves Daoust
    Dec 16 '18 at 20:03














1












1








1





$begingroup$



$$ int :frac{2y^3}{sqrt{1-y^2}}mathrm dy $$




I substitute $$ y = sin t $$ then $$ 2 int :frac{sin^3 t cdot cos t}{cos t};mathrm dt = 2 int sin^2 t;mathrm d cos t $$
What should I do next, and how do I convert back to the original variable $y$?






calculus integration substitution







share|cite|improve this question











$endgroup$





$$ int :frac{2y^3}{sqrt{1-y^2}}mathrm dy $$




I substitute $$ y = sin t $$ then $$ 2 int :frac{sin^3 t cdot cos t}{cos t};mathrm dt = 2 int sin^2 t;mathrm d cos t $$
What should I do next, and how do I convert back to the original variable $y$?






calculus integration substitution




calculus integration substitution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 19:07









Algebear

707419




707419










asked Dec 16 '18 at 18:40









Dmitry SokolovDmitry Sokolov

526




526












  • $begingroup$
    Why not $$sin t=y$$
    $endgroup$
    – lab bhattacharjee
    Dec 16 '18 at 18:41






  • 1




    $begingroup$
    Another way: $$sqrt{1-y^2}=u$$
    $endgroup$
    – lab bhattacharjee
    Dec 16 '18 at 18:41






  • 1




    $begingroup$
    $sin^2t=1-cos^2t$.
    $endgroup$
    – Yves Daoust
    Dec 16 '18 at 20:03


















  • $begingroup$
    Why not $$sin t=y$$
    $endgroup$
    – lab bhattacharjee
    Dec 16 '18 at 18:41






  • 1




    $begingroup$
    Another way: $$sqrt{1-y^2}=u$$
    $endgroup$
    – lab bhattacharjee
    Dec 16 '18 at 18:41






  • 1




    $begingroup$
    $sin^2t=1-cos^2t$.
    $endgroup$
    – Yves Daoust
    Dec 16 '18 at 20:03
















$begingroup$
Why not $$sin t=y$$
$endgroup$
– lab bhattacharjee
Dec 16 '18 at 18:41




$begingroup$
Why not $$sin t=y$$
$endgroup$
– lab bhattacharjee
Dec 16 '18 at 18:41




1




1




$begingroup$
Another way: $$sqrt{1-y^2}=u$$
$endgroup$
– lab bhattacharjee
Dec 16 '18 at 18:41




$begingroup$
Another way: $$sqrt{1-y^2}=u$$
$endgroup$
– lab bhattacharjee
Dec 16 '18 at 18:41




1




1




$begingroup$
$sin^2t=1-cos^2t$.
$endgroup$
– Yves Daoust
Dec 16 '18 at 20:03




$begingroup$
$sin^2t=1-cos^2t$.
$endgroup$
– Yves Daoust
Dec 16 '18 at 20:03










4 Answers
4






active

oldest

votes


















3












$begingroup$

$$I=2intfrac{x^3}{sqrt{1-x^2}}mathrm{d}x$$
$x=sin u$:
$$I=2intsin^3u mathrm{d}u$$
$$I=2int(1-cos^2u)sin u mathrm{d}u$$
$w=cos u$:
$$I=2int(w^2-1)mathrm{d}w$$
$$I=2int w^2mathrm{d}w-2int mathrm{d}w$$
$$I=frac23w^3-2w$$
$$I=frac23cos^3u-2cos u$$
$$I=frac23(1-x^2)^{3/2}-2(1-x^2)^{1/2}+C$$
$$I=-frac23(2+x^2)sqrt{1-x^2}+C$$






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    This type of problem is much easier if you do a $u-$sub by picking $u=1-y^2$ and $du=-2ydy,$ then



    $$int frac{2y^3dy}{sqrt{1-y^2}}=-intfrac{(1-u)}{sqrt{u}}du$$
    and this becomes a simple power rule computation afterward.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      You've established the following relationship.



      $$int{frac{2y^3}{sqrt{1-y^2}}dy}=int{2sin^2tspace dt}+C$$
      Where $C$ is a fixed constant.
      You can evaluate the second integral using the rule $$cos(2alpha)=1-2sin^2(alpha)$$
      Then use that $y=sin t to t=sin^{-1}(y)$ to change the integral back.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Though I find the substitution $u=sqrt{1-y^2}$ leading to the most elegant solution of this integral, you are asking for a help with the integral obtained through the substitution $y=sin t.$



        $$begin{aligned}int :frac{sin^3 t cdot cos t}{cos t};mathrm dt& = int sin^2 t;mathrm d cos t\&=int (1-cos ^2 t);mathrm d cos t,end{aligned}$$ which is easy to finish.

        Then use $:t=arcsin y:$ and $;cos (arcsin y)=sqrt{1-y^2}.$






        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%2f3042974%2fintegral-by-trigonometric-substitution-int-frac2y3-sqrt1-y2-mathr%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          $$I=2intfrac{x^3}{sqrt{1-x^2}}mathrm{d}x$$
          $x=sin u$:
          $$I=2intsin^3u mathrm{d}u$$
          $$I=2int(1-cos^2u)sin u mathrm{d}u$$
          $w=cos u$:
          $$I=2int(w^2-1)mathrm{d}w$$
          $$I=2int w^2mathrm{d}w-2int mathrm{d}w$$
          $$I=frac23w^3-2w$$
          $$I=frac23cos^3u-2cos u$$
          $$I=frac23(1-x^2)^{3/2}-2(1-x^2)^{1/2}+C$$
          $$I=-frac23(2+x^2)sqrt{1-x^2}+C$$






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            $$I=2intfrac{x^3}{sqrt{1-x^2}}mathrm{d}x$$
            $x=sin u$:
            $$I=2intsin^3u mathrm{d}u$$
            $$I=2int(1-cos^2u)sin u mathrm{d}u$$
            $w=cos u$:
            $$I=2int(w^2-1)mathrm{d}w$$
            $$I=2int w^2mathrm{d}w-2int mathrm{d}w$$
            $$I=frac23w^3-2w$$
            $$I=frac23cos^3u-2cos u$$
            $$I=frac23(1-x^2)^{3/2}-2(1-x^2)^{1/2}+C$$
            $$I=-frac23(2+x^2)sqrt{1-x^2}+C$$






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              $$I=2intfrac{x^3}{sqrt{1-x^2}}mathrm{d}x$$
              $x=sin u$:
              $$I=2intsin^3u mathrm{d}u$$
              $$I=2int(1-cos^2u)sin u mathrm{d}u$$
              $w=cos u$:
              $$I=2int(w^2-1)mathrm{d}w$$
              $$I=2int w^2mathrm{d}w-2int mathrm{d}w$$
              $$I=frac23w^3-2w$$
              $$I=frac23cos^3u-2cos u$$
              $$I=frac23(1-x^2)^{3/2}-2(1-x^2)^{1/2}+C$$
              $$I=-frac23(2+x^2)sqrt{1-x^2}+C$$






              share|cite|improve this answer









              $endgroup$



              $$I=2intfrac{x^3}{sqrt{1-x^2}}mathrm{d}x$$
              $x=sin u$:
              $$I=2intsin^3u mathrm{d}u$$
              $$I=2int(1-cos^2u)sin u mathrm{d}u$$
              $w=cos u$:
              $$I=2int(w^2-1)mathrm{d}w$$
              $$I=2int w^2mathrm{d}w-2int mathrm{d}w$$
              $$I=frac23w^3-2w$$
              $$I=frac23cos^3u-2cos u$$
              $$I=frac23(1-x^2)^{3/2}-2(1-x^2)^{1/2}+C$$
              $$I=-frac23(2+x^2)sqrt{1-x^2}+C$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 16 '18 at 19:56









              clathratusclathratus

              4,9601338




              4,9601338























                  4












                  $begingroup$

                  This type of problem is much easier if you do a $u-$sub by picking $u=1-y^2$ and $du=-2ydy,$ then



                  $$int frac{2y^3dy}{sqrt{1-y^2}}=-intfrac{(1-u)}{sqrt{u}}du$$
                  and this becomes a simple power rule computation afterward.






                  share|cite|improve this answer











                  $endgroup$


















                    4












                    $begingroup$

                    This type of problem is much easier if you do a $u-$sub by picking $u=1-y^2$ and $du=-2ydy,$ then



                    $$int frac{2y^3dy}{sqrt{1-y^2}}=-intfrac{(1-u)}{sqrt{u}}du$$
                    and this becomes a simple power rule computation afterward.






                    share|cite|improve this answer











                    $endgroup$
















                      4












                      4








                      4





                      $begingroup$

                      This type of problem is much easier if you do a $u-$sub by picking $u=1-y^2$ and $du=-2ydy,$ then



                      $$int frac{2y^3dy}{sqrt{1-y^2}}=-intfrac{(1-u)}{sqrt{u}}du$$
                      and this becomes a simple power rule computation afterward.






                      share|cite|improve this answer











                      $endgroup$



                      This type of problem is much easier if you do a $u-$sub by picking $u=1-y^2$ and $du=-2ydy,$ then



                      $$int frac{2y^3dy}{sqrt{1-y^2}}=-intfrac{(1-u)}{sqrt{u}}du$$
                      and this becomes a simple power rule computation afterward.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 16 '18 at 20:11

























                      answered Dec 16 '18 at 18:44









                      ChickenmancerChickenmancer

                      3,324724




                      3,324724























                          1












                          $begingroup$

                          You've established the following relationship.



                          $$int{frac{2y^3}{sqrt{1-y^2}}dy}=int{2sin^2tspace dt}+C$$
                          Where $C$ is a fixed constant.
                          You can evaluate the second integral using the rule $$cos(2alpha)=1-2sin^2(alpha)$$
                          Then use that $y=sin t to t=sin^{-1}(y)$ to change the integral back.






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            You've established the following relationship.



                            $$int{frac{2y^3}{sqrt{1-y^2}}dy}=int{2sin^2tspace dt}+C$$
                            Where $C$ is a fixed constant.
                            You can evaluate the second integral using the rule $$cos(2alpha)=1-2sin^2(alpha)$$
                            Then use that $y=sin t to t=sin^{-1}(y)$ to change the integral back.






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              You've established the following relationship.



                              $$int{frac{2y^3}{sqrt{1-y^2}}dy}=int{2sin^2tspace dt}+C$$
                              Where $C$ is a fixed constant.
                              You can evaluate the second integral using the rule $$cos(2alpha)=1-2sin^2(alpha)$$
                              Then use that $y=sin t to t=sin^{-1}(y)$ to change the integral back.






                              share|cite|improve this answer









                              $endgroup$



                              You've established the following relationship.



                              $$int{frac{2y^3}{sqrt{1-y^2}}dy}=int{2sin^2tspace dt}+C$$
                              Where $C$ is a fixed constant.
                              You can evaluate the second integral using the rule $$cos(2alpha)=1-2sin^2(alpha)$$
                              Then use that $y=sin t to t=sin^{-1}(y)$ to change the integral back.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 16 '18 at 18:53









                              Rhys HughesRhys Hughes

                              6,9901630




                              6,9901630























                                  0












                                  $begingroup$

                                  Though I find the substitution $u=sqrt{1-y^2}$ leading to the most elegant solution of this integral, you are asking for a help with the integral obtained through the substitution $y=sin t.$



                                  $$begin{aligned}int :frac{sin^3 t cdot cos t}{cos t};mathrm dt& = int sin^2 t;mathrm d cos t\&=int (1-cos ^2 t);mathrm d cos t,end{aligned}$$ which is easy to finish.

                                  Then use $:t=arcsin y:$ and $;cos (arcsin y)=sqrt{1-y^2}.$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Though I find the substitution $u=sqrt{1-y^2}$ leading to the most elegant solution of this integral, you are asking for a help with the integral obtained through the substitution $y=sin t.$



                                    $$begin{aligned}int :frac{sin^3 t cdot cos t}{cos t};mathrm dt& = int sin^2 t;mathrm d cos t\&=int (1-cos ^2 t);mathrm d cos t,end{aligned}$$ which is easy to finish.

                                    Then use $:t=arcsin y:$ and $;cos (arcsin y)=sqrt{1-y^2}.$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Though I find the substitution $u=sqrt{1-y^2}$ leading to the most elegant solution of this integral, you are asking for a help with the integral obtained through the substitution $y=sin t.$



                                      $$begin{aligned}int :frac{sin^3 t cdot cos t}{cos t};mathrm dt& = int sin^2 t;mathrm d cos t\&=int (1-cos ^2 t);mathrm d cos t,end{aligned}$$ which is easy to finish.

                                      Then use $:t=arcsin y:$ and $;cos (arcsin y)=sqrt{1-y^2}.$






                                      share|cite|improve this answer









                                      $endgroup$



                                      Though I find the substitution $u=sqrt{1-y^2}$ leading to the most elegant solution of this integral, you are asking for a help with the integral obtained through the substitution $y=sin t.$



                                      $$begin{aligned}int :frac{sin^3 t cdot cos t}{cos t};mathrm dt& = int sin^2 t;mathrm d cos t\&=int (1-cos ^2 t);mathrm d cos t,end{aligned}$$ which is easy to finish.

                                      Then use $:t=arcsin y:$ and $;cos (arcsin y)=sqrt{1-y^2}.$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 16 '18 at 20:42









                                      user376343user376343

                                      3,9584829




                                      3,9584829






























                                          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%2f3042974%2fintegral-by-trigonometric-substitution-int-frac2y3-sqrt1-y2-mathr%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

                                          Plaza Victoria

                                          Image
                                          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                                          Read more

                                          Brian Clough

                                          Image
                                          Brian Clough Datos personales Nombre completo Brian Howard Clough Apodo(s) Old Big 'Ead, Cloughie [ 1 ] ​ Nacimiento Middlesbrough, Inglaterra, Reino Unido 21 de marzo de 1935 Nacionalidad(es) Fallecimiento Derby, Reino Unido 20 de septiembre de 2004 (69 años) Altura 1,78 [ 1 ] ​ metros Carrera Deporte Fútbol Debut deportivo 1955 (Como jugador) 1965 (Como entrenador) (Middlesbrough F. C. (Como jugador) Hartlepool United F. C. (Como entrenador) ) Posición Delantero Goles en clubes 269 [ 2 ] ​ Retirada deportiva 1964 (Como jugador) 1993 (Como entrenador) (Sunderland A. F. C. (Como jugador) Nottingham Forest (Como entrenador) ) Carrera internacional Part. 2 [editar datos en Wikidata] Brian Howard Clough , OBE [ 3 ] ​ (Middlesbrough, Inglaterra, Reino Unido, 21 de marzo de 1935 – Derby, Inglaterra, Reino Unido, 20 de septiembre de 2004) fue un jugador y entrenador británico de fútbol. Es conocido por...
                                          Read more

                                          Cáceres

                                          Image
                                          Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
                                          Read more