Evaluate $I_n = int_0^{pi / 2} sin n theta cos theta ,dtheta$ by integrating by parts twice












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By integrating by parts twice, show that $I_n$, as defined below for integers $n > 1$, has the value shown.




$$I_n = int_0^{pi / 2} sin n theta cos theta ,dtheta = frac{n-sin(frac{pi n}{2})}{n^2 -1}$$




I can do this using the formula $$sin A cos B = frac{1}{2}[sin(A-B)+sin(A+B)] ,$$ but when I try using integration by parts I get stuck in a loop of integrating the same thing over and over.










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    Please format your question a bit more; there are apparently some unreadable characters.
    – T. Bongers
    Nov 18 at 19:04










  • Can you describe a little more the choices that lead to getting "stuck in a loop"?
    – Travis
    Nov 18 at 19:18










  • @Travis let theta=x so it's quicker for me to type... so, after the first IBP i get sin(x)sin(nx) - Intergral [n(sin(x)cos(nx))]. when i use IBP on this integral i get sin(x)sin(nx) - Integral [cos(x)sin(nx)] which is what i originally tried to integrate ..
    – Taylor
    Nov 18 at 19:48










  • @T.Bongers i did try my best. i'm new to this.
    – Taylor
    Nov 18 at 19:48










  • I think that there are some non-Unicode symbols being included in the equation (to the right of the second equality).
    – T. Bongers
    Nov 18 at 19:49
















0














By integrating by parts twice, show that $I_n$, as defined below for integers $n > 1$, has the value shown.




$$I_n = int_0^{pi / 2} sin n theta cos theta ,dtheta = frac{n-sin(frac{pi n}{2})}{n^2 -1}$$




I can do this using the formula $$sin A cos B = frac{1}{2}[sin(A-B)+sin(A+B)] ,$$ but when I try using integration by parts I get stuck in a loop of integrating the same thing over and over.










share|cite|improve this question




















  • 1




    Please format your question a bit more; there are apparently some unreadable characters.
    – T. Bongers
    Nov 18 at 19:04










  • Can you describe a little more the choices that lead to getting "stuck in a loop"?
    – Travis
    Nov 18 at 19:18










  • @Travis let theta=x so it's quicker for me to type... so, after the first IBP i get sin(x)sin(nx) - Intergral [n(sin(x)cos(nx))]. when i use IBP on this integral i get sin(x)sin(nx) - Integral [cos(x)sin(nx)] which is what i originally tried to integrate ..
    – Taylor
    Nov 18 at 19:48










  • @T.Bongers i did try my best. i'm new to this.
    – Taylor
    Nov 18 at 19:48










  • I think that there are some non-Unicode symbols being included in the equation (to the right of the second equality).
    – T. Bongers
    Nov 18 at 19:49














0












0








0







By integrating by parts twice, show that $I_n$, as defined below for integers $n > 1$, has the value shown.




$$I_n = int_0^{pi / 2} sin n theta cos theta ,dtheta = frac{n-sin(frac{pi n}{2})}{n^2 -1}$$




I can do this using the formula $$sin A cos B = frac{1}{2}[sin(A-B)+sin(A+B)] ,$$ but when I try using integration by parts I get stuck in a loop of integrating the same thing over and over.










share|cite|improve this question















By integrating by parts twice, show that $I_n$, as defined below for integers $n > 1$, has the value shown.




$$I_n = int_0^{pi / 2} sin n theta cos theta ,dtheta = frac{n-sin(frac{pi n}{2})}{n^2 -1}$$




I can do this using the formula $$sin A cos B = frac{1}{2}[sin(A-B)+sin(A+B)] ,$$ but when I try using integration by parts I get stuck in a loop of integrating the same thing over and over.







calculus integration definite-integrals trigonometric-integrals






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edited Nov 25 at 8:51









Martin Sleziak

44.6k7115270




44.6k7115270










asked Nov 18 at 19:03









Taylor

42




42








  • 1




    Please format your question a bit more; there are apparently some unreadable characters.
    – T. Bongers
    Nov 18 at 19:04










  • Can you describe a little more the choices that lead to getting "stuck in a loop"?
    – Travis
    Nov 18 at 19:18










  • @Travis let theta=x so it's quicker for me to type... so, after the first IBP i get sin(x)sin(nx) - Intergral [n(sin(x)cos(nx))]. when i use IBP on this integral i get sin(x)sin(nx) - Integral [cos(x)sin(nx)] which is what i originally tried to integrate ..
    – Taylor
    Nov 18 at 19:48










  • @T.Bongers i did try my best. i'm new to this.
    – Taylor
    Nov 18 at 19:48










  • I think that there are some non-Unicode symbols being included in the equation (to the right of the second equality).
    – T. Bongers
    Nov 18 at 19:49














  • 1




    Please format your question a bit more; there are apparently some unreadable characters.
    – T. Bongers
    Nov 18 at 19:04










  • Can you describe a little more the choices that lead to getting "stuck in a loop"?
    – Travis
    Nov 18 at 19:18










  • @Travis let theta=x so it's quicker for me to type... so, after the first IBP i get sin(x)sin(nx) - Intergral [n(sin(x)cos(nx))]. when i use IBP on this integral i get sin(x)sin(nx) - Integral [cos(x)sin(nx)] which is what i originally tried to integrate ..
    – Taylor
    Nov 18 at 19:48










  • @T.Bongers i did try my best. i'm new to this.
    – Taylor
    Nov 18 at 19:48










  • I think that there are some non-Unicode symbols being included in the equation (to the right of the second equality).
    – T. Bongers
    Nov 18 at 19:49








1




1




Please format your question a bit more; there are apparently some unreadable characters.
– T. Bongers
Nov 18 at 19:04




Please format your question a bit more; there are apparently some unreadable characters.
– T. Bongers
Nov 18 at 19:04












Can you describe a little more the choices that lead to getting "stuck in a loop"?
– Travis
Nov 18 at 19:18




Can you describe a little more the choices that lead to getting "stuck in a loop"?
– Travis
Nov 18 at 19:18












@Travis let theta=x so it's quicker for me to type... so, after the first IBP i get sin(x)sin(nx) - Intergral [n(sin(x)cos(nx))]. when i use IBP on this integral i get sin(x)sin(nx) - Integral [cos(x)sin(nx)] which is what i originally tried to integrate ..
– Taylor
Nov 18 at 19:48




@Travis let theta=x so it's quicker for me to type... so, after the first IBP i get sin(x)sin(nx) - Intergral [n(sin(x)cos(nx))]. when i use IBP on this integral i get sin(x)sin(nx) - Integral [cos(x)sin(nx)] which is what i originally tried to integrate ..
– Taylor
Nov 18 at 19:48












@T.Bongers i did try my best. i'm new to this.
– Taylor
Nov 18 at 19:48




@T.Bongers i did try my best. i'm new to this.
– Taylor
Nov 18 at 19:48












I think that there are some non-Unicode symbols being included in the equation (to the right of the second equality).
– T. Bongers
Nov 18 at 19:49




I think that there are some non-Unicode symbols being included in the equation (to the right of the second equality).
– T. Bongers
Nov 18 at 19:49










3 Answers
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Hint Following what you've done already, integrating by parts with $u = sin n theta$, $dv = cos theta ,dtheta$ gives
begin{multline}color{#df0000}{I_n} = underbrace{sin n theta}_u , underbrace{sin theta}_v vert_0^{pi / 2} - int_0^{pi / 2} underbrace{sin theta}_v , underbrace{cos ntheta , dtheta}_{du} = sin frac{pi n}{2} - n color{#1f1fff}{J_n}, \ color{#1f1fff}{J_n := int_0^{pi / 2} cos n theta sin theta , dtheta} .end{multline}



We now apply integration by parts to the integral $color{#3f3fff}{J_n}$ with $p = cos n theta$, $dq = sin theta ,dtheta$:
$$color{#3f3fff}{J_n} = cos n theta (-cos theta)vert_0^{pi / 2} - int_0^{pi / 2} underbrace{-cos theta}_q cdot underbrace{- n sin n theta ,dtheta}_{dp} = 1 - n color{#df0000}{I_n} .$$




Substituting to eliminate $color{#3f3fff}{J_n}$ gives $color{#df0000}{I_n} = sin frac{pi n}{2} - n (1 - n color{#df0000}{I_n})$, and rearranging to solve for $color{#df0000}{I_n}$ gives the claimed identity: $$color{#df0000}{boxed{I_n = frac{n - sin frac{pi n}{2}}{n^2 - 1}}} .$$ Notice that for $n equiv 0, 2 pmod 4$ this simplifies to $frac{n}{n^2 - 1}$, for $n equiv 1 pmod 4$ to $frac{1}{n + 1}$, and for $n equiv 3 pmod 4$ to $frac{1}{n - 1}$.







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    0














    $$cos{(x)} sin{left( n xright) }=frac{sin{left( left( n+1right) xright) }+sin{left( left( n-1right) xright) }}{2}$$
    $$int{left. cos{(x)} sin{left( n xright) }dxright.}=-frac{cos{left( left( n+1right) xright) }}{2 left( n+1right) }-frac{cos{left( left( n-1right) xright) }}{2 left( n-1right) }$$
    Then
    $$int_{0}^{frac{pi}{2}}{left. cos{(x)} sin{left( n xright) }dxright.}\=
    frac{n}{n^2-1}-frac{left( n-1right) cos{left( frac{{pi} n+{pi} }{2}right) }+left( n+1right) cos{left( frac{{pi} n-{pi} }{2}right) }}{2n^2-2}\=
    frac{n}{{{n}^{2}}-1}-frac{sin{left( frac{{pi} n}{2}right) }}{{{n}^{2}}-1}$$






    share|cite|improve this answer





























      0














      Hint:



      There's an escape from the infinite loop. If you observe closely, you will see that after two iterations you get to something like



      $$I=b+aI$$



      where the constants $a,b$ are computable.



      You naturally conclude



      $$I=frac{b}{1-a}.$$






      $$intsin nthetacostheta,dtheta=sin nthetasintheta-nintcos nthetasintheta,dtheta\=sin nthetasintheta+ncos nthetacostheta+n^2intsin nthetacostheta,dtheta.$$







      share|cite|improve this answer























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        3 Answers
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        3 Answers
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        active

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        active

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        2














        Hint Following what you've done already, integrating by parts with $u = sin n theta$, $dv = cos theta ,dtheta$ gives
        begin{multline}color{#df0000}{I_n} = underbrace{sin n theta}_u , underbrace{sin theta}_v vert_0^{pi / 2} - int_0^{pi / 2} underbrace{sin theta}_v , underbrace{cos ntheta , dtheta}_{du} = sin frac{pi n}{2} - n color{#1f1fff}{J_n}, \ color{#1f1fff}{J_n := int_0^{pi / 2} cos n theta sin theta , dtheta} .end{multline}



        We now apply integration by parts to the integral $color{#3f3fff}{J_n}$ with $p = cos n theta$, $dq = sin theta ,dtheta$:
        $$color{#3f3fff}{J_n} = cos n theta (-cos theta)vert_0^{pi / 2} - int_0^{pi / 2} underbrace{-cos theta}_q cdot underbrace{- n sin n theta ,dtheta}_{dp} = 1 - n color{#df0000}{I_n} .$$




        Substituting to eliminate $color{#3f3fff}{J_n}$ gives $color{#df0000}{I_n} = sin frac{pi n}{2} - n (1 - n color{#df0000}{I_n})$, and rearranging to solve for $color{#df0000}{I_n}$ gives the claimed identity: $$color{#df0000}{boxed{I_n = frac{n - sin frac{pi n}{2}}{n^2 - 1}}} .$$ Notice that for $n equiv 0, 2 pmod 4$ this simplifies to $frac{n}{n^2 - 1}$, for $n equiv 1 pmod 4$ to $frac{1}{n + 1}$, and for $n equiv 3 pmod 4$ to $frac{1}{n - 1}$.







        share|cite|improve this answer




























          2














          Hint Following what you've done already, integrating by parts with $u = sin n theta$, $dv = cos theta ,dtheta$ gives
          begin{multline}color{#df0000}{I_n} = underbrace{sin n theta}_u , underbrace{sin theta}_v vert_0^{pi / 2} - int_0^{pi / 2} underbrace{sin theta}_v , underbrace{cos ntheta , dtheta}_{du} = sin frac{pi n}{2} - n color{#1f1fff}{J_n}, \ color{#1f1fff}{J_n := int_0^{pi / 2} cos n theta sin theta , dtheta} .end{multline}



          We now apply integration by parts to the integral $color{#3f3fff}{J_n}$ with $p = cos n theta$, $dq = sin theta ,dtheta$:
          $$color{#3f3fff}{J_n} = cos n theta (-cos theta)vert_0^{pi / 2} - int_0^{pi / 2} underbrace{-cos theta}_q cdot underbrace{- n sin n theta ,dtheta}_{dp} = 1 - n color{#df0000}{I_n} .$$




          Substituting to eliminate $color{#3f3fff}{J_n}$ gives $color{#df0000}{I_n} = sin frac{pi n}{2} - n (1 - n color{#df0000}{I_n})$, and rearranging to solve for $color{#df0000}{I_n}$ gives the claimed identity: $$color{#df0000}{boxed{I_n = frac{n - sin frac{pi n}{2}}{n^2 - 1}}} .$$ Notice that for $n equiv 0, 2 pmod 4$ this simplifies to $frac{n}{n^2 - 1}$, for $n equiv 1 pmod 4$ to $frac{1}{n + 1}$, and for $n equiv 3 pmod 4$ to $frac{1}{n - 1}$.







          share|cite|improve this answer


























            2












            2








            2






            Hint Following what you've done already, integrating by parts with $u = sin n theta$, $dv = cos theta ,dtheta$ gives
            begin{multline}color{#df0000}{I_n} = underbrace{sin n theta}_u , underbrace{sin theta}_v vert_0^{pi / 2} - int_0^{pi / 2} underbrace{sin theta}_v , underbrace{cos ntheta , dtheta}_{du} = sin frac{pi n}{2} - n color{#1f1fff}{J_n}, \ color{#1f1fff}{J_n := int_0^{pi / 2} cos n theta sin theta , dtheta} .end{multline}



            We now apply integration by parts to the integral $color{#3f3fff}{J_n}$ with $p = cos n theta$, $dq = sin theta ,dtheta$:
            $$color{#3f3fff}{J_n} = cos n theta (-cos theta)vert_0^{pi / 2} - int_0^{pi / 2} underbrace{-cos theta}_q cdot underbrace{- n sin n theta ,dtheta}_{dp} = 1 - n color{#df0000}{I_n} .$$




            Substituting to eliminate $color{#3f3fff}{J_n}$ gives $color{#df0000}{I_n} = sin frac{pi n}{2} - n (1 - n color{#df0000}{I_n})$, and rearranging to solve for $color{#df0000}{I_n}$ gives the claimed identity: $$color{#df0000}{boxed{I_n = frac{n - sin frac{pi n}{2}}{n^2 - 1}}} .$$ Notice that for $n equiv 0, 2 pmod 4$ this simplifies to $frac{n}{n^2 - 1}$, for $n equiv 1 pmod 4$ to $frac{1}{n + 1}$, and for $n equiv 3 pmod 4$ to $frac{1}{n - 1}$.







            share|cite|improve this answer














            Hint Following what you've done already, integrating by parts with $u = sin n theta$, $dv = cos theta ,dtheta$ gives
            begin{multline}color{#df0000}{I_n} = underbrace{sin n theta}_u , underbrace{sin theta}_v vert_0^{pi / 2} - int_0^{pi / 2} underbrace{sin theta}_v , underbrace{cos ntheta , dtheta}_{du} = sin frac{pi n}{2} - n color{#1f1fff}{J_n}, \ color{#1f1fff}{J_n := int_0^{pi / 2} cos n theta sin theta , dtheta} .end{multline}



            We now apply integration by parts to the integral $color{#3f3fff}{J_n}$ with $p = cos n theta$, $dq = sin theta ,dtheta$:
            $$color{#3f3fff}{J_n} = cos n theta (-cos theta)vert_0^{pi / 2} - int_0^{pi / 2} underbrace{-cos theta}_q cdot underbrace{- n sin n theta ,dtheta}_{dp} = 1 - n color{#df0000}{I_n} .$$




            Substituting to eliminate $color{#3f3fff}{J_n}$ gives $color{#df0000}{I_n} = sin frac{pi n}{2} - n (1 - n color{#df0000}{I_n})$, and rearranging to solve for $color{#df0000}{I_n}$ gives the claimed identity: $$color{#df0000}{boxed{I_n = frac{n - sin frac{pi n}{2}}{n^2 - 1}}} .$$ Notice that for $n equiv 0, 2 pmod 4$ this simplifies to $frac{n}{n^2 - 1}$, for $n equiv 1 pmod 4$ to $frac{1}{n + 1}$, and for $n equiv 3 pmod 4$ to $frac{1}{n - 1}$.








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            edited Nov 25 at 7:23

























            answered Nov 18 at 21:27









            Travis

            59.4k767145




            59.4k767145























                0














                $$cos{(x)} sin{left( n xright) }=frac{sin{left( left( n+1right) xright) }+sin{left( left( n-1right) xright) }}{2}$$
                $$int{left. cos{(x)} sin{left( n xright) }dxright.}=-frac{cos{left( left( n+1right) xright) }}{2 left( n+1right) }-frac{cos{left( left( n-1right) xright) }}{2 left( n-1right) }$$
                Then
                $$int_{0}^{frac{pi}{2}}{left. cos{(x)} sin{left( n xright) }dxright.}\=
                frac{n}{n^2-1}-frac{left( n-1right) cos{left( frac{{pi} n+{pi} }{2}right) }+left( n+1right) cos{left( frac{{pi} n-{pi} }{2}right) }}{2n^2-2}\=
                frac{n}{{{n}^{2}}-1}-frac{sin{left( frac{{pi} n}{2}right) }}{{{n}^{2}}-1}$$






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                  0














                  $$cos{(x)} sin{left( n xright) }=frac{sin{left( left( n+1right) xright) }+sin{left( left( n-1right) xright) }}{2}$$
                  $$int{left. cos{(x)} sin{left( n xright) }dxright.}=-frac{cos{left( left( n+1right) xright) }}{2 left( n+1right) }-frac{cos{left( left( n-1right) xright) }}{2 left( n-1right) }$$
                  Then
                  $$int_{0}^{frac{pi}{2}}{left. cos{(x)} sin{left( n xright) }dxright.}\=
                  frac{n}{n^2-1}-frac{left( n-1right) cos{left( frac{{pi} n+{pi} }{2}right) }+left( n+1right) cos{left( frac{{pi} n-{pi} }{2}right) }}{2n^2-2}\=
                  frac{n}{{{n}^{2}}-1}-frac{sin{left( frac{{pi} n}{2}right) }}{{{n}^{2}}-1}$$






                  share|cite|improve this answer
























                    0












                    0








                    0






                    $$cos{(x)} sin{left( n xright) }=frac{sin{left( left( n+1right) xright) }+sin{left( left( n-1right) xright) }}{2}$$
                    $$int{left. cos{(x)} sin{left( n xright) }dxright.}=-frac{cos{left( left( n+1right) xright) }}{2 left( n+1right) }-frac{cos{left( left( n-1right) xright) }}{2 left( n-1right) }$$
                    Then
                    $$int_{0}^{frac{pi}{2}}{left. cos{(x)} sin{left( n xright) }dxright.}\=
                    frac{n}{n^2-1}-frac{left( n-1right) cos{left( frac{{pi} n+{pi} }{2}right) }+left( n+1right) cos{left( frac{{pi} n-{pi} }{2}right) }}{2n^2-2}\=
                    frac{n}{{{n}^{2}}-1}-frac{sin{left( frac{{pi} n}{2}right) }}{{{n}^{2}}-1}$$






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                    $$cos{(x)} sin{left( n xright) }=frac{sin{left( left( n+1right) xright) }+sin{left( left( n-1right) xright) }}{2}$$
                    $$int{left. cos{(x)} sin{left( n xright) }dxright.}=-frac{cos{left( left( n+1right) xright) }}{2 left( n+1right) }-frac{cos{left( left( n-1right) xright) }}{2 left( n-1right) }$$
                    Then
                    $$int_{0}^{frac{pi}{2}}{left. cos{(x)} sin{left( n xright) }dxright.}\=
                    frac{n}{n^2-1}-frac{left( n-1right) cos{left( frac{{pi} n+{pi} }{2}right) }+left( n+1right) cos{left( frac{{pi} n-{pi} }{2}right) }}{2n^2-2}\=
                    frac{n}{{{n}^{2}}-1}-frac{sin{left( frac{{pi} n}{2}right) }}{{{n}^{2}}-1}$$







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                    answered Nov 18 at 20:11









                    Aleksas Domarkas

                    8276




                    8276























                        0














                        Hint:



                        There's an escape from the infinite loop. If you observe closely, you will see that after two iterations you get to something like



                        $$I=b+aI$$



                        where the constants $a,b$ are computable.



                        You naturally conclude



                        $$I=frac{b}{1-a}.$$






                        $$intsin nthetacostheta,dtheta=sin nthetasintheta-nintcos nthetasintheta,dtheta\=sin nthetasintheta+ncos nthetacostheta+n^2intsin nthetacostheta,dtheta.$$







                        share|cite|improve this answer




























                          0














                          Hint:



                          There's an escape from the infinite loop. If you observe closely, you will see that after two iterations you get to something like



                          $$I=b+aI$$



                          where the constants $a,b$ are computable.



                          You naturally conclude



                          $$I=frac{b}{1-a}.$$






                          $$intsin nthetacostheta,dtheta=sin nthetasintheta-nintcos nthetasintheta,dtheta\=sin nthetasintheta+ncos nthetacostheta+n^2intsin nthetacostheta,dtheta.$$







                          share|cite|improve this answer


























                            0












                            0








                            0






                            Hint:



                            There's an escape from the infinite loop. If you observe closely, you will see that after two iterations you get to something like



                            $$I=b+aI$$



                            where the constants $a,b$ are computable.



                            You naturally conclude



                            $$I=frac{b}{1-a}.$$






                            $$intsin nthetacostheta,dtheta=sin nthetasintheta-nintcos nthetasintheta,dtheta\=sin nthetasintheta+ncos nthetacostheta+n^2intsin nthetacostheta,dtheta.$$







                            share|cite|improve this answer














                            Hint:



                            There's an escape from the infinite loop. If you observe closely, you will see that after two iterations you get to something like



                            $$I=b+aI$$



                            where the constants $a,b$ are computable.



                            You naturally conclude



                            $$I=frac{b}{1-a}.$$






                            $$intsin nthetacostheta,dtheta=sin nthetasintheta-nintcos nthetasintheta,dtheta\=sin nthetasintheta+ncos nthetacostheta+n^2intsin nthetacostheta,dtheta.$$








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                            edited Nov 25 at 9:11

























                            answered Nov 25 at 9:02









                            Yves Daoust

                            124k671221




                            124k671221






























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