A Difficult Definite Integral $int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t.$












4












$begingroup$


Problem



Evaluate $$int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t.$$



Comment



It's very complicated to compute the integral applying normal method. I obtain the result resorting to the skillful formula




$$int_0^{2pi}xf(cos x){rm d}x=piint_0^{2pi}f(sin x){rm d}x,$$
where $f(x) in C[-1,1].$




begin{align*}
require{begingroup}
begingroup
newcommand{dd}{;{rm d}}int_0^{2pi} (t-sin t)(1-cos t)^2 dd t
&= int_0^{2pi} t(1-cos t)^2 dd t - int_0^{2pi} sin t(1-cos t)^2 dd t \
&= piint_0^{2pi} (1-sin t)^2 dd t - int_0^{2pi} (1-cos t)^2 dd (1-cos t) \
&= piint_0^{2pi} left(frac32-frac12cos2t-2sin tright) dd t - left[frac13(1-cos t)^3right]_0^{2pi}\
&= pileft[frac32t-frac14sin2t+2cos tright]_0^{2pi}\
&= 3pi^2
endgroup
end{align*}



But any other solution?










share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Problem



    Evaluate $$int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t.$$



    Comment



    It's very complicated to compute the integral applying normal method. I obtain the result resorting to the skillful formula




    $$int_0^{2pi}xf(cos x){rm d}x=piint_0^{2pi}f(sin x){rm d}x,$$
    where $f(x) in C[-1,1].$




    begin{align*}
    require{begingroup}
    begingroup
    newcommand{dd}{;{rm d}}int_0^{2pi} (t-sin t)(1-cos t)^2 dd t
    &= int_0^{2pi} t(1-cos t)^2 dd t - int_0^{2pi} sin t(1-cos t)^2 dd t \
    &= piint_0^{2pi} (1-sin t)^2 dd t - int_0^{2pi} (1-cos t)^2 dd (1-cos t) \
    &= piint_0^{2pi} left(frac32-frac12cos2t-2sin tright) dd t - left[frac13(1-cos t)^3right]_0^{2pi}\
    &= pileft[frac32t-frac14sin2t+2cos tright]_0^{2pi}\
    &= 3pi^2
    endgroup
    end{align*}



    But any other solution?










    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      Problem



      Evaluate $$int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t.$$



      Comment



      It's very complicated to compute the integral applying normal method. I obtain the result resorting to the skillful formula




      $$int_0^{2pi}xf(cos x){rm d}x=piint_0^{2pi}f(sin x){rm d}x,$$
      where $f(x) in C[-1,1].$




      begin{align*}
      require{begingroup}
      begingroup
      newcommand{dd}{;{rm d}}int_0^{2pi} (t-sin t)(1-cos t)^2 dd t
      &= int_0^{2pi} t(1-cos t)^2 dd t - int_0^{2pi} sin t(1-cos t)^2 dd t \
      &= piint_0^{2pi} (1-sin t)^2 dd t - int_0^{2pi} (1-cos t)^2 dd (1-cos t) \
      &= piint_0^{2pi} left(frac32-frac12cos2t-2sin tright) dd t - left[frac13(1-cos t)^3right]_0^{2pi}\
      &= pileft[frac32t-frac14sin2t+2cos tright]_0^{2pi}\
      &= 3pi^2
      endgroup
      end{align*}



      But any other solution?










      share|cite|improve this question











      $endgroup$




      Problem



      Evaluate $$int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t.$$



      Comment



      It's very complicated to compute the integral applying normal method. I obtain the result resorting to the skillful formula




      $$int_0^{2pi}xf(cos x){rm d}x=piint_0^{2pi}f(sin x){rm d}x,$$
      where $f(x) in C[-1,1].$




      begin{align*}
      require{begingroup}
      begingroup
      newcommand{dd}{;{rm d}}int_0^{2pi} (t-sin t)(1-cos t)^2 dd t
      &= int_0^{2pi} t(1-cos t)^2 dd t - int_0^{2pi} sin t(1-cos t)^2 dd t \
      &= piint_0^{2pi} (1-sin t)^2 dd t - int_0^{2pi} (1-cos t)^2 dd (1-cos t) \
      &= piint_0^{2pi} left(frac32-frac12cos2t-2sin tright) dd t - left[frac13(1-cos t)^3right]_0^{2pi}\
      &= pileft[frac32t-frac14sin2t+2cos tright]_0^{2pi}\
      &= 3pi^2
      endgroup
      end{align*}



      But any other solution?







      integration definite-integrals trigonometric-integrals






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      edited Dec 28 '18 at 16:31









      Martin Sleziak

      44.7k8117272




      44.7k8117272










      asked Dec 28 '18 at 13:48









      mengdie1982mengdie1982

      4,826618




      4,826618






















          4 Answers
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          6












          $begingroup$

          With substitution $t=pi+x$ we see that
          $$I=int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_{-pi}^{pi}(pi+x+sin x)(1+cos x)^2{rm d}x.$$
          the part $(x+sin x)(1+cos x)^2$ is an odd function so it's integral over $[-pi,pi]$ is zero, then
          $$I=piint_{-pi}^{pi}(1+cos x)^2{rm d}x=piint_{-pi}^{pi}dfrac32+2cos x+dfrac12cos2x{rm d}x=color{blue}{3pi^2}$$






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            Notice that your integral is in fact an area integral of the function $(x,y) mapsto x$ over one arch of a cycloid given by
            $$begin{cases} x = t - sin t \ y = 1-cos tend{cases}$$
            Indeed, if we assume that the cycloid is explicitly given by $y = f(x)$, we have
            begin{align}
            int_{text{cycloid}} x ,dA &= int_0^{2pi} int_0^{f(x)} x,dy,dx\
            &= int_0^{2pi} xf(x),dx \
            &= begin{bmatrix} x = t-sin t \ f(x) = 1-cos t \ dx = (1-cos t),dtend{bmatrix}\
            &= int_0^{2pi} (t-sin t)(1-cos t)^2,dt
            end{align}



            Now recall the formula for the $x$-coordinate of the centroid:
            $$frac1A int_{text{cycloid}} x ,dA = xtext{-coordinate of the centroid} = pi$$



            since centroid is clearly at $x = pi$ by symmetry.



            Using the same metrod as above area of the cycloid is $$A = int_{text{cycloid}}dA = int_0^{2pi} (1-cos t)^2 ,dt = 3pi$$



            this integral being a lot easier than the original one.



            It follows $$int_{text{cycloid}} x ,dA = Api = 3pi^2$$






            share|cite|improve this answer











            $endgroup$





















              1












              $begingroup$

              The integral
              $$
              int_0^{2pi}sin t(1-cos t)^2,dt=Bigl[frac{(1-cos t)^3}{3}Bigr]_0^{2pi}=0
              $$

              is immediate. Thus we can concentrate on
              $$
              int_0^{2pi}t(1-cos t)^2,dt=[dots t=2u dots]=
              16int_0^{pi}usin^4u,du=int_0^pi u(e^{iu}-e^{-iu})^4,du
              $$

              For integer $a$, we have
              $$
              int_0^pi ue^{2iau},du=Bigl[frac{ue^{2iau}}{2ia}Bigr]_0^pi-frac{1}{2ia}int_0^pi e^{2iau},du=frac{pi e^{2iapi}}{2ia}+frac{1}{4a^2}Bigl[e^{2iau}Bigr]_0^pi=frac{pi}{2ia}
              $$

              Since $(e^{iu}-e^{-iu})^4=e^{4iu}-4e^{2iu}+6-4e^{-2iu}+e^{-4iu}$ we see that
              $$
              int_0^pi usin^4u,du=int_0^pi 6u,du=3pi^2
              $$






              share|cite|improve this answer









              $endgroup$





















                1












                $begingroup$

                begin{align}
                &int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_0^{2pi}t+t{cos }^2 t-2tcos t-sin t -sin t{cos }^2 t+sin 2t{rm d}t\
                &=left[frac{1}{2}t^2+frac{1}{2}t^2+frac{1}{4}tsin 2t-frac{1}{4}t^2+frac{1}{8}cos 2t-2tsin t-2cos t+cos t+frac{1}{3}{cos }^3t-frac{1}{2}cos 2tright]_0^{2pi} \
                &=left[frac{3}{4}t^2+frac{1}{4}tsin 2t-frac{3}{8}cos 2t-2tsin t-cos t+frac{1}{3}{cos }^3 tright]_0^{2pi}\
                &=3pi ^2
                end{align}






                share|cite|improve this answer











                $endgroup$













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                  4 Answers
                  4






                  active

                  oldest

                  votes








                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  6












                  $begingroup$

                  With substitution $t=pi+x$ we see that
                  $$I=int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_{-pi}^{pi}(pi+x+sin x)(1+cos x)^2{rm d}x.$$
                  the part $(x+sin x)(1+cos x)^2$ is an odd function so it's integral over $[-pi,pi]$ is zero, then
                  $$I=piint_{-pi}^{pi}(1+cos x)^2{rm d}x=piint_{-pi}^{pi}dfrac32+2cos x+dfrac12cos2x{rm d}x=color{blue}{3pi^2}$$






                  share|cite|improve this answer









                  $endgroup$


















                    6












                    $begingroup$

                    With substitution $t=pi+x$ we see that
                    $$I=int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_{-pi}^{pi}(pi+x+sin x)(1+cos x)^2{rm d}x.$$
                    the part $(x+sin x)(1+cos x)^2$ is an odd function so it's integral over $[-pi,pi]$ is zero, then
                    $$I=piint_{-pi}^{pi}(1+cos x)^2{rm d}x=piint_{-pi}^{pi}dfrac32+2cos x+dfrac12cos2x{rm d}x=color{blue}{3pi^2}$$






                    share|cite|improve this answer









                    $endgroup$
















                      6












                      6








                      6





                      $begingroup$

                      With substitution $t=pi+x$ we see that
                      $$I=int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_{-pi}^{pi}(pi+x+sin x)(1+cos x)^2{rm d}x.$$
                      the part $(x+sin x)(1+cos x)^2$ is an odd function so it's integral over $[-pi,pi]$ is zero, then
                      $$I=piint_{-pi}^{pi}(1+cos x)^2{rm d}x=piint_{-pi}^{pi}dfrac32+2cos x+dfrac12cos2x{rm d}x=color{blue}{3pi^2}$$






                      share|cite|improve this answer









                      $endgroup$



                      With substitution $t=pi+x$ we see that
                      $$I=int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_{-pi}^{pi}(pi+x+sin x)(1+cos x)^2{rm d}x.$$
                      the part $(x+sin x)(1+cos x)^2$ is an odd function so it's integral over $[-pi,pi]$ is zero, then
                      $$I=piint_{-pi}^{pi}(1+cos x)^2{rm d}x=piint_{-pi}^{pi}dfrac32+2cos x+dfrac12cos2x{rm d}x=color{blue}{3pi^2}$$







                      share|cite|improve this answer












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                      share|cite|improve this answer










                      answered Dec 28 '18 at 16:26









                      NosratiNosrati

                      26.5k62354




                      26.5k62354























                          4












                          $begingroup$

                          Notice that your integral is in fact an area integral of the function $(x,y) mapsto x$ over one arch of a cycloid given by
                          $$begin{cases} x = t - sin t \ y = 1-cos tend{cases}$$
                          Indeed, if we assume that the cycloid is explicitly given by $y = f(x)$, we have
                          begin{align}
                          int_{text{cycloid}} x ,dA &= int_0^{2pi} int_0^{f(x)} x,dy,dx\
                          &= int_0^{2pi} xf(x),dx \
                          &= begin{bmatrix} x = t-sin t \ f(x) = 1-cos t \ dx = (1-cos t),dtend{bmatrix}\
                          &= int_0^{2pi} (t-sin t)(1-cos t)^2,dt
                          end{align}



                          Now recall the formula for the $x$-coordinate of the centroid:
                          $$frac1A int_{text{cycloid}} x ,dA = xtext{-coordinate of the centroid} = pi$$



                          since centroid is clearly at $x = pi$ by symmetry.



                          Using the same metrod as above area of the cycloid is $$A = int_{text{cycloid}}dA = int_0^{2pi} (1-cos t)^2 ,dt = 3pi$$



                          this integral being a lot easier than the original one.



                          It follows $$int_{text{cycloid}} x ,dA = Api = 3pi^2$$






                          share|cite|improve this answer











                          $endgroup$


















                            4












                            $begingroup$

                            Notice that your integral is in fact an area integral of the function $(x,y) mapsto x$ over one arch of a cycloid given by
                            $$begin{cases} x = t - sin t \ y = 1-cos tend{cases}$$
                            Indeed, if we assume that the cycloid is explicitly given by $y = f(x)$, we have
                            begin{align}
                            int_{text{cycloid}} x ,dA &= int_0^{2pi} int_0^{f(x)} x,dy,dx\
                            &= int_0^{2pi} xf(x),dx \
                            &= begin{bmatrix} x = t-sin t \ f(x) = 1-cos t \ dx = (1-cos t),dtend{bmatrix}\
                            &= int_0^{2pi} (t-sin t)(1-cos t)^2,dt
                            end{align}



                            Now recall the formula for the $x$-coordinate of the centroid:
                            $$frac1A int_{text{cycloid}} x ,dA = xtext{-coordinate of the centroid} = pi$$



                            since centroid is clearly at $x = pi$ by symmetry.



                            Using the same metrod as above area of the cycloid is $$A = int_{text{cycloid}}dA = int_0^{2pi} (1-cos t)^2 ,dt = 3pi$$



                            this integral being a lot easier than the original one.



                            It follows $$int_{text{cycloid}} x ,dA = Api = 3pi^2$$






                            share|cite|improve this answer











                            $endgroup$
















                              4












                              4








                              4





                              $begingroup$

                              Notice that your integral is in fact an area integral of the function $(x,y) mapsto x$ over one arch of a cycloid given by
                              $$begin{cases} x = t - sin t \ y = 1-cos tend{cases}$$
                              Indeed, if we assume that the cycloid is explicitly given by $y = f(x)$, we have
                              begin{align}
                              int_{text{cycloid}} x ,dA &= int_0^{2pi} int_0^{f(x)} x,dy,dx\
                              &= int_0^{2pi} xf(x),dx \
                              &= begin{bmatrix} x = t-sin t \ f(x) = 1-cos t \ dx = (1-cos t),dtend{bmatrix}\
                              &= int_0^{2pi} (t-sin t)(1-cos t)^2,dt
                              end{align}



                              Now recall the formula for the $x$-coordinate of the centroid:
                              $$frac1A int_{text{cycloid}} x ,dA = xtext{-coordinate of the centroid} = pi$$



                              since centroid is clearly at $x = pi$ by symmetry.



                              Using the same metrod as above area of the cycloid is $$A = int_{text{cycloid}}dA = int_0^{2pi} (1-cos t)^2 ,dt = 3pi$$



                              this integral being a lot easier than the original one.



                              It follows $$int_{text{cycloid}} x ,dA = Api = 3pi^2$$






                              share|cite|improve this answer











                              $endgroup$



                              Notice that your integral is in fact an area integral of the function $(x,y) mapsto x$ over one arch of a cycloid given by
                              $$begin{cases} x = t - sin t \ y = 1-cos tend{cases}$$
                              Indeed, if we assume that the cycloid is explicitly given by $y = f(x)$, we have
                              begin{align}
                              int_{text{cycloid}} x ,dA &= int_0^{2pi} int_0^{f(x)} x,dy,dx\
                              &= int_0^{2pi} xf(x),dx \
                              &= begin{bmatrix} x = t-sin t \ f(x) = 1-cos t \ dx = (1-cos t),dtend{bmatrix}\
                              &= int_0^{2pi} (t-sin t)(1-cos t)^2,dt
                              end{align}



                              Now recall the formula for the $x$-coordinate of the centroid:
                              $$frac1A int_{text{cycloid}} x ,dA = xtext{-coordinate of the centroid} = pi$$



                              since centroid is clearly at $x = pi$ by symmetry.



                              Using the same metrod as above area of the cycloid is $$A = int_{text{cycloid}}dA = int_0^{2pi} (1-cos t)^2 ,dt = 3pi$$



                              this integral being a lot easier than the original one.



                              It follows $$int_{text{cycloid}} x ,dA = Api = 3pi^2$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Dec 28 '18 at 14:59

























                              answered Dec 28 '18 at 14:50









                              mechanodroidmechanodroid

                              27.1k62446




                              27.1k62446























                                  1












                                  $begingroup$

                                  The integral
                                  $$
                                  int_0^{2pi}sin t(1-cos t)^2,dt=Bigl[frac{(1-cos t)^3}{3}Bigr]_0^{2pi}=0
                                  $$

                                  is immediate. Thus we can concentrate on
                                  $$
                                  int_0^{2pi}t(1-cos t)^2,dt=[dots t=2u dots]=
                                  16int_0^{pi}usin^4u,du=int_0^pi u(e^{iu}-e^{-iu})^4,du
                                  $$

                                  For integer $a$, we have
                                  $$
                                  int_0^pi ue^{2iau},du=Bigl[frac{ue^{2iau}}{2ia}Bigr]_0^pi-frac{1}{2ia}int_0^pi e^{2iau},du=frac{pi e^{2iapi}}{2ia}+frac{1}{4a^2}Bigl[e^{2iau}Bigr]_0^pi=frac{pi}{2ia}
                                  $$

                                  Since $(e^{iu}-e^{-iu})^4=e^{4iu}-4e^{2iu}+6-4e^{-2iu}+e^{-4iu}$ we see that
                                  $$
                                  int_0^pi usin^4u,du=int_0^pi 6u,du=3pi^2
                                  $$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    The integral
                                    $$
                                    int_0^{2pi}sin t(1-cos t)^2,dt=Bigl[frac{(1-cos t)^3}{3}Bigr]_0^{2pi}=0
                                    $$

                                    is immediate. Thus we can concentrate on
                                    $$
                                    int_0^{2pi}t(1-cos t)^2,dt=[dots t=2u dots]=
                                    16int_0^{pi}usin^4u,du=int_0^pi u(e^{iu}-e^{-iu})^4,du
                                    $$

                                    For integer $a$, we have
                                    $$
                                    int_0^pi ue^{2iau},du=Bigl[frac{ue^{2iau}}{2ia}Bigr]_0^pi-frac{1}{2ia}int_0^pi e^{2iau},du=frac{pi e^{2iapi}}{2ia}+frac{1}{4a^2}Bigl[e^{2iau}Bigr]_0^pi=frac{pi}{2ia}
                                    $$

                                    Since $(e^{iu}-e^{-iu})^4=e^{4iu}-4e^{2iu}+6-4e^{-2iu}+e^{-4iu}$ we see that
                                    $$
                                    int_0^pi usin^4u,du=int_0^pi 6u,du=3pi^2
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      The integral
                                      $$
                                      int_0^{2pi}sin t(1-cos t)^2,dt=Bigl[frac{(1-cos t)^3}{3}Bigr]_0^{2pi}=0
                                      $$

                                      is immediate. Thus we can concentrate on
                                      $$
                                      int_0^{2pi}t(1-cos t)^2,dt=[dots t=2u dots]=
                                      16int_0^{pi}usin^4u,du=int_0^pi u(e^{iu}-e^{-iu})^4,du
                                      $$

                                      For integer $a$, we have
                                      $$
                                      int_0^pi ue^{2iau},du=Bigl[frac{ue^{2iau}}{2ia}Bigr]_0^pi-frac{1}{2ia}int_0^pi e^{2iau},du=frac{pi e^{2iapi}}{2ia}+frac{1}{4a^2}Bigl[e^{2iau}Bigr]_0^pi=frac{pi}{2ia}
                                      $$

                                      Since $(e^{iu}-e^{-iu})^4=e^{4iu}-4e^{2iu}+6-4e^{-2iu}+e^{-4iu}$ we see that
                                      $$
                                      int_0^pi usin^4u,du=int_0^pi 6u,du=3pi^2
                                      $$






                                      share|cite|improve this answer









                                      $endgroup$



                                      The integral
                                      $$
                                      int_0^{2pi}sin t(1-cos t)^2,dt=Bigl[frac{(1-cos t)^3}{3}Bigr]_0^{2pi}=0
                                      $$

                                      is immediate. Thus we can concentrate on
                                      $$
                                      int_0^{2pi}t(1-cos t)^2,dt=[dots t=2u dots]=
                                      16int_0^{pi}usin^4u,du=int_0^pi u(e^{iu}-e^{-iu})^4,du
                                      $$

                                      For integer $a$, we have
                                      $$
                                      int_0^pi ue^{2iau},du=Bigl[frac{ue^{2iau}}{2ia}Bigr]_0^pi-frac{1}{2ia}int_0^pi e^{2iau},du=frac{pi e^{2iapi}}{2ia}+frac{1}{4a^2}Bigl[e^{2iau}Bigr]_0^pi=frac{pi}{2ia}
                                      $$

                                      Since $(e^{iu}-e^{-iu})^4=e^{4iu}-4e^{2iu}+6-4e^{-2iu}+e^{-4iu}$ we see that
                                      $$
                                      int_0^pi usin^4u,du=int_0^pi 6u,du=3pi^2
                                      $$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 28 '18 at 15:39









                                      egregegreg

                                      180k1485202




                                      180k1485202























                                          1












                                          $begingroup$

                                          begin{align}
                                          &int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_0^{2pi}t+t{cos }^2 t-2tcos t-sin t -sin t{cos }^2 t+sin 2t{rm d}t\
                                          &=left[frac{1}{2}t^2+frac{1}{2}t^2+frac{1}{4}tsin 2t-frac{1}{4}t^2+frac{1}{8}cos 2t-2tsin t-2cos t+cos t+frac{1}{3}{cos }^3t-frac{1}{2}cos 2tright]_0^{2pi} \
                                          &=left[frac{3}{4}t^2+frac{1}{4}tsin 2t-frac{3}{8}cos 2t-2tsin t-cos t+frac{1}{3}{cos }^3 tright]_0^{2pi}\
                                          &=3pi ^2
                                          end{align}






                                          share|cite|improve this answer











                                          $endgroup$


















                                            1












                                            $begingroup$

                                            begin{align}
                                            &int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_0^{2pi}t+t{cos }^2 t-2tcos t-sin t -sin t{cos }^2 t+sin 2t{rm d}t\
                                            &=left[frac{1}{2}t^2+frac{1}{2}t^2+frac{1}{4}tsin 2t-frac{1}{4}t^2+frac{1}{8}cos 2t-2tsin t-2cos t+cos t+frac{1}{3}{cos }^3t-frac{1}{2}cos 2tright]_0^{2pi} \
                                            &=left[frac{3}{4}t^2+frac{1}{4}tsin 2t-frac{3}{8}cos 2t-2tsin t-cos t+frac{1}{3}{cos }^3 tright]_0^{2pi}\
                                            &=3pi ^2
                                            end{align}






                                            share|cite|improve this answer











                                            $endgroup$
















                                              1












                                              1








                                              1





                                              $begingroup$

                                              begin{align}
                                              &int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_0^{2pi}t+t{cos }^2 t-2tcos t-sin t -sin t{cos }^2 t+sin 2t{rm d}t\
                                              &=left[frac{1}{2}t^2+frac{1}{2}t^2+frac{1}{4}tsin 2t-frac{1}{4}t^2+frac{1}{8}cos 2t-2tsin t-2cos t+cos t+frac{1}{3}{cos }^3t-frac{1}{2}cos 2tright]_0^{2pi} \
                                              &=left[frac{3}{4}t^2+frac{1}{4}tsin 2t-frac{3}{8}cos 2t-2tsin t-cos t+frac{1}{3}{cos }^3 tright]_0^{2pi}\
                                              &=3pi ^2
                                              end{align}






                                              share|cite|improve this answer











                                              $endgroup$



                                              begin{align}
                                              &int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_0^{2pi}t+t{cos }^2 t-2tcos t-sin t -sin t{cos }^2 t+sin 2t{rm d}t\
                                              &=left[frac{1}{2}t^2+frac{1}{2}t^2+frac{1}{4}tsin 2t-frac{1}{4}t^2+frac{1}{8}cos 2t-2tsin t-2cos t+cos t+frac{1}{3}{cos }^3t-frac{1}{2}cos 2tright]_0^{2pi} \
                                              &=left[frac{3}{4}t^2+frac{1}{4}tsin 2t-frac{3}{8}cos 2t-2tsin t-cos t+frac{1}{3}{cos }^3 tright]_0^{2pi}\
                                              &=3pi ^2
                                              end{align}







                                              share|cite|improve this answer














                                              share|cite|improve this answer



                                              share|cite|improve this answer








                                              edited Dec 29 '18 at 5:05









                                              DavidG

                                              1,964620




                                              1,964620










                                              answered Dec 28 '18 at 14:12









                                              yavaryavar

                                              823




                                              823






























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