How to evaluate $int_pi/6^pi/3fracsqrt[3]sin xsqrt[3]sin x+sqrt[3]cos x$?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












$$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos x$$



I have tried it using transformation but its getting lengthy..it must be short question. please help










share|cite|improve this question



























    up vote
    0
    down vote

    favorite












    $$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos x$$



    I have tried it using transformation but its getting lengthy..it must be short question. please help










    share|cite|improve this question

























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      $$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos x$$



      I have tried it using transformation but its getting lengthy..it must be short question. please help










      share|cite|improve this question















      $$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos x$$



      I have tried it using transformation but its getting lengthy..it must be short question. please help







      integration definite-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 5 '16 at 10:46









      StubbornAtom

      4,11711136




      4,11711136










      asked Nov 5 '16 at 10:20









      Rajnesh Singh

      31




      31




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          Hint use $f (x)=f (a+b-x ) $ where a,b are lower ,upper limits . Then everything is straightforward






          share|cite|improve this answer



























            up vote
            0
            down vote













            Using the hint given by Archis Welankar:
            $$int_a^b f(x) dx=int_a^bf(a+b-x)dx;\
            int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin (pi/2-x) sqrt[3]sin (pi/2-x) + sqrt[3]cos (pi/2-x)dx Rightarrow \
            underbraceint_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx_A=underbraceint_pi/6^pi/3 frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx_B Rightarrow \
            A-B=0.$$
            Also:
            $$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin x+sqrt[3]cos x-sqrt[3]cos x sqrt[3]sin x + sqrt[3]cos xdx=\
            int_pi/6^pi/3 1-frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx Rightarrow \
            A+B=fracpi6.$$
            Hence:
            $$A=fracpi12.$$






            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',
              convertImagesToLinks: true,
              noModals: false,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              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%2f2000317%2fhow-to-evaluate-int-pi-6-pi-3-frac-sqrt3-sin-x-sqrt3-sin-x-s%23new-answer', 'question_page');

              );

              Post as a guest






























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote



              accepted










              Hint use $f (x)=f (a+b-x ) $ where a,b are lower ,upper limits . Then everything is straightforward






              share|cite|improve this answer
























                up vote
                1
                down vote



                accepted










                Hint use $f (x)=f (a+b-x ) $ where a,b are lower ,upper limits . Then everything is straightforward






                share|cite|improve this answer






















                  up vote
                  1
                  down vote



                  accepted







                  up vote
                  1
                  down vote



                  accepted






                  Hint use $f (x)=f (a+b-x ) $ where a,b are lower ,upper limits . Then everything is straightforward






                  share|cite|improve this answer












                  Hint use $f (x)=f (a+b-x ) $ where a,b are lower ,upper limits . Then everything is straightforward







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 5 '16 at 10:23









                  Archis Welankar

                  11.9k41239




                  11.9k41239




















                      up vote
                      0
                      down vote













                      Using the hint given by Archis Welankar:
                      $$int_a^b f(x) dx=int_a^bf(a+b-x)dx;\
                      int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin (pi/2-x) sqrt[3]sin (pi/2-x) + sqrt[3]cos (pi/2-x)dx Rightarrow \
                      underbraceint_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx_A=underbraceint_pi/6^pi/3 frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx_B Rightarrow \
                      A-B=0.$$
                      Also:
                      $$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin x+sqrt[3]cos x-sqrt[3]cos x sqrt[3]sin x + sqrt[3]cos xdx=\
                      int_pi/6^pi/3 1-frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx Rightarrow \
                      A+B=fracpi6.$$
                      Hence:
                      $$A=fracpi12.$$






                      share|cite|improve this answer
























                        up vote
                        0
                        down vote













                        Using the hint given by Archis Welankar:
                        $$int_a^b f(x) dx=int_a^bf(a+b-x)dx;\
                        int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin (pi/2-x) sqrt[3]sin (pi/2-x) + sqrt[3]cos (pi/2-x)dx Rightarrow \
                        underbraceint_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx_A=underbraceint_pi/6^pi/3 frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx_B Rightarrow \
                        A-B=0.$$
                        Also:
                        $$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin x+sqrt[3]cos x-sqrt[3]cos x sqrt[3]sin x + sqrt[3]cos xdx=\
                        int_pi/6^pi/3 1-frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx Rightarrow \
                        A+B=fracpi6.$$
                        Hence:
                        $$A=fracpi12.$$






                        share|cite|improve this answer






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Using the hint given by Archis Welankar:
                          $$int_a^b f(x) dx=int_a^bf(a+b-x)dx;\
                          int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin (pi/2-x) sqrt[3]sin (pi/2-x) + sqrt[3]cos (pi/2-x)dx Rightarrow \
                          underbraceint_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx_A=underbraceint_pi/6^pi/3 frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx_B Rightarrow \
                          A-B=0.$$
                          Also:
                          $$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin x+sqrt[3]cos x-sqrt[3]cos x sqrt[3]sin x + sqrt[3]cos xdx=\
                          int_pi/6^pi/3 1-frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx Rightarrow \
                          A+B=fracpi6.$$
                          Hence:
                          $$A=fracpi12.$$






                          share|cite|improve this answer












                          Using the hint given by Archis Welankar:
                          $$int_a^b f(x) dx=int_a^bf(a+b-x)dx;\
                          int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin (pi/2-x) sqrt[3]sin (pi/2-x) + sqrt[3]cos (pi/2-x)dx Rightarrow \
                          underbraceint_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx_A=underbraceint_pi/6^pi/3 frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx_B Rightarrow \
                          A-B=0.$$
                          Also:
                          $$int_pi/6^pi/3 frac sqrt[3]sin x sqrt[3]sin x + sqrt[3]cos xdx=int_pi/6^pi/3 frac sqrt[3]sin x+sqrt[3]cos x-sqrt[3]cos x sqrt[3]sin x + sqrt[3]cos xdx=\
                          int_pi/6^pi/3 1-frac sqrt[3]cos x sqrt[3]cos x + sqrt[3]sin xdx Rightarrow \
                          A+B=fracpi6.$$
                          Hence:
                          $$A=fracpi12.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Sep 9 at 12:38









                          farruhota

                          15.6k2734




                          15.6k2734



























                               

                              draft saved


                              draft discarded















































                               


                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2000317%2fhow-to-evaluate-int-pi-6-pi-3-frac-sqrt3-sin-x-sqrt3-sin-x-s%23new-answer', 'question_page');

                              );

                              Post as a guest













































































                              這個網誌中的熱門文章

                              Why am i infinitely getting the same tweet with the Twitter Search API?

                              Is there any way to eliminate the singular point to solve this integral by hand or by approximations?

                              Strongly p-embedded subgroups and p-Sylow subgroups.