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
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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












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                            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












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                                    share|cite|improve this answer










                                    answered 2 hours ago









                                    yavar

                                    643




                                    643






























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                                        広島県道265号伴広島線

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                                        一般県道 広島県道265号伴広島線 路線延長 km 起点 広島県広島市安佐南区伴東7丁目 終点 広島県広島市中区榎町 接続する 主な道路 （記法） 広島県道38号広島豊平線 広島県道71号広島湯来線 山陽自動車道五日市IC 国道183号（ 国道261号重複） ■テンプレート（■ノート ■使い方） ■PJ道路 広島県道265号伴広島線 （ひろしまけんどう265ごう ともひろしません）は広島県広島市安佐南区から広島県広島市中区を結ぶ一般県道である。 目次 1 路線データ 2 通過する自治体 3 見どころ 4 接続道路 4.1 本線 4.2 旧道 5 別名 6 関連項目 路線データ 本線 起点：広島市安佐南区伴東7丁目・大原下橋南詰交差点（広島県道38号広島豊平線（本線・旧道）交点） 終点：広島市中区榎町・十日市交差点（国道183号（国道261号重用）交点） 総延長： 旧道 起点：広島市安佐南区伴東4丁目・新細坂交差点（広島県道38号広島豊平線交点） 終点：広島市安佐南区伴中央4丁目・寺組橋(南)交差点（広島県道265号伴広島線（本線）交点） 総延長： 通過する自治体 広島県 広島市（安佐南区、佐伯区、西区、中区） 見どころ 広島高速交通広島新交通1号線（アストラムライン） 広島広域公園 広島広域公園陸上競技場（広島ビッグアーチ） 己斐峠 己斐橋（太田川放水路） 天満橋（天満川） 接続道路 本線 広島県道38号広島豊平線（本線・旧道）（広島市安佐南区伴東7丁目・大原下橋南詰交差点〔起点〕） 広島県道71号広島湯来線（本線）（広島市安佐南区伴中央2丁目・伴交番入口交差点 - 広島市安佐南区伴中央4丁目・寺組橋(南)交差点間で重用） 広島県道71号広島湯来線（旧道）（広島県道265号伴広島線（旧道）重用）（広島市安佐南区伴中央4丁目・寺組橋(南)交差点） 広島県道71号広島湯来線（広島市安佐南区伴中央4丁目・寺組橋(南)交差点 - 広島市佐伯区五日市町石内・五月が丘(南)交差点間で重用）
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                                        Setup Asymptote in Texstudio

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                                        0 Good time of the day! I am trying to compile the following file: documentclass[12pt]{article} usepackage[inline]{asymptote} begin{document} Here is a venn diagram produced with Asymptote, drawn to width 4cm: defA{A} defB{V{B}} begin{figure} begin{center} begin{asy} size(4cm,0); pen colour1=red; pen colour2=green; pair z0=(0,0); pair z1=(-1,0); pair z2=(1,0); real r=1.5; path c1=circle(z1,r); path c2=circle(z2,r); fill(c1,colour1); fill(c2,colour2); picture intersection=new picture; fill(intersection,c1,colour1+colour2); clip(intersection,c2); add(intersection); draw(c1); draw(c2); //draw("$A$",box,z1)
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