A Difficult Definite Integral












3














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 $f(x) in C[-1,1].$




enter image description hereBut any other solution?






definite-integrals







share|cite|improve this question



























    3














    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 $f(x) in C[-1,1].$




    enter image description hereBut any other solution?






    definite-integrals







    share|cite|improve this question

























      3












      3








      3







      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 $f(x) in C[-1,1].$




      enter image description hereBut any other solution?






      definite-integrals







      share|cite|improve this question













      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 $f(x) in C[-1,1].$




      enter image description hereBut any other solution?






      definite-integrals




      definite-integrals






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 hours ago









      mengdie1982

      4,775618




      4,775618






















          3 Answers
          3






          active

          oldest

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          2














          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































            1














            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





























              0














              $int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_0^{2pi}t+t{cos }^2 t-2tcos t-sin t -sin t{cos }^2 t+sin 2t{rm d}t=frac{1}{2}t^2+frac{1}{2}t^2+frac{1}{4}tsin 2t-frac{1}{4}t^2+frac{1}{8}cos 2t-2tsin t-2cos t+cos t+frac{1}{3}{cos }^3t-frac{1}{2}cos 2t=frac{3}{4}t^2+frac{1}{4}tsin 2t-frac{3}{8}cos 2t-2tsin t-cos t+frac{1}{3}{cos }^3 t=3pi ^2$






              share|cite|improve this answer





















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                3 Answers
                3






                active

                oldest

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                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

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                2














                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




























                  2














                  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


























                    2












                    2








                    2






                    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














                    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 1 hour ago

























                    answered 1 hour ago









                    mechanodroid

                    25.9k62245




                    25.9k62245























                        1














                        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


























                          1














                          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
























                            1












                            1








                            1






                            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












                            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 39 mins ago









                            egreg

                            178k1484200




                            178k1484200























                                0














                                $int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_0^{2pi}t+t{cos }^2 t-2tcos t-sin t -sin t{cos }^2 t+sin 2t{rm d}t=frac{1}{2}t^2+frac{1}{2}t^2+frac{1}{4}tsin 2t-frac{1}{4}t^2+frac{1}{8}cos 2t-2tsin t-2cos t+cos t+frac{1}{3}{cos }^3t-frac{1}{2}cos 2t=frac{3}{4}t^2+frac{1}{4}tsin 2t-frac{3}{8}cos 2t-2tsin t-cos t+frac{1}{3}{cos }^3 t=3pi ^2$






                                share|cite|improve this answer


























                                  0














                                  $int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_0^{2pi}t+t{cos }^2 t-2tcos t-sin t -sin t{cos }^2 t+sin 2t{rm d}t=frac{1}{2}t^2+frac{1}{2}t^2+frac{1}{4}tsin 2t-frac{1}{4}t^2+frac{1}{8}cos 2t-2tsin t-2cos t+cos t+frac{1}{3}{cos }^3t-frac{1}{2}cos 2t=frac{3}{4}t^2+frac{1}{4}tsin 2t-frac{3}{8}cos 2t-2tsin t-cos t+frac{1}{3}{cos }^3 t=3pi ^2$






                                  share|cite|improve this answer
























                                    0












                                    0








                                    0






                                    $int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_0^{2pi}t+t{cos }^2 t-2tcos t-sin t -sin t{cos }^2 t+sin 2t{rm d}t=frac{1}{2}t^2+frac{1}{2}t^2+frac{1}{4}tsin 2t-frac{1}{4}t^2+frac{1}{8}cos 2t-2tsin t-2cos t+cos t+frac{1}{3}{cos }^3t-frac{1}{2}cos 2t=frac{3}{4}t^2+frac{1}{4}tsin 2t-frac{3}{8}cos 2t-2tsin t-cos t+frac{1}{3}{cos }^3 t=3pi ^2$






                                    share|cite|improve this answer












                                    $int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_0^{2pi}t+t{cos }^2 t-2tcos t-sin t -sin t{cos }^2 t+sin 2t{rm d}t=frac{1}{2}t^2+frac{1}{2}t^2+frac{1}{4}tsin 2t-frac{1}{4}t^2+frac{1}{8}cos 2t-2tsin t-2cos t+cos t+frac{1}{3}{cos }^3t-frac{1}{2}cos 2t=frac{3}{4}t^2+frac{1}{4}tsin 2t-frac{3}{8}cos 2t-2tsin t-cos t+frac{1}{3}{cos }^3 t=3pi ^2$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 2 hours ago









                                    yavar

                                    643




                                    643






























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