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

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



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










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



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










    share|cite|improve this question

























      up vote
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      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






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






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            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:
            $$A=fracpi12.$$






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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

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              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












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                  answered Nov 5 '16 at 10:23









                  Archis Welankar

                  11.9k41239




                  11.9k41239




















                      up vote
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                      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:
                          $$A=fracpi12.$$







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                          answered Sep 9 at 12:38









                          farruhota

                          15.6k2734




                          15.6k2734



























                               

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