finding value of integral












3












$begingroup$


$$int_0^{pi/2} (cos x)^{2011} sin (2013 x) dx $$



The value is given to be $frac{1}{2012}$. I am while solving using by parts, but one term is remaining like $frac{1}{2}$, then $frac{1}{3},$ and so on. Kindly help.



what i am stuck at is following.



$$I(m,n) =int_0^{pi/2} (cos x)^{m} sin (n x) dx $$
$$I(m,n)= frac {1}{m+n} + frac{m}{m+n}I(m-1,n-1)$$
after deriving this i am not able to proceed further






calculus definite-integrals







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    why downvoting?
    $endgroup$
    – maveric
    Dec 15 '18 at 8:45






  • 3




    $begingroup$
    In this case, reasons for the down- and closevotes would be helpful. I do not see one.
    $endgroup$
    – Peter
    Dec 15 '18 at 9:20










  • $begingroup$
    What's meant by stuck? You can not derive that or don't know how to utilize that
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 9:37










  • $begingroup$
    i have reached till this step. i am able to proceed further to answer which is 1/2011
    $endgroup$
    – maveric
    Dec 15 '18 at 9:38










  • $begingroup$
    This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
    $endgroup$
    – Jean Marie
    Dec 15 '18 at 9:51
















3












$begingroup$


$$int_0^{pi/2} (cos x)^{2011} sin (2013 x) dx $$



The value is given to be $frac{1}{2012}$. I am while solving using by parts, but one term is remaining like $frac{1}{2}$, then $frac{1}{3},$ and so on. Kindly help.



what i am stuck at is following.



$$I(m,n) =int_0^{pi/2} (cos x)^{m} sin (n x) dx $$
$$I(m,n)= frac {1}{m+n} + frac{m}{m+n}I(m-1,n-1)$$
after deriving this i am not able to proceed further






calculus definite-integrals







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    why downvoting?
    $endgroup$
    – maveric
    Dec 15 '18 at 8:45






  • 3




    $begingroup$
    In this case, reasons for the down- and closevotes would be helpful. I do not see one.
    $endgroup$
    – Peter
    Dec 15 '18 at 9:20










  • $begingroup$
    What's meant by stuck? You can not derive that or don't know how to utilize that
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 9:37










  • $begingroup$
    i have reached till this step. i am able to proceed further to answer which is 1/2011
    $endgroup$
    – maveric
    Dec 15 '18 at 9:38










  • $begingroup$
    This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
    $endgroup$
    – Jean Marie
    Dec 15 '18 at 9:51














3












3








3


1



$begingroup$


$$int_0^{pi/2} (cos x)^{2011} sin (2013 x) dx $$



The value is given to be $frac{1}{2012}$. I am while solving using by parts, but one term is remaining like $frac{1}{2}$, then $frac{1}{3},$ and so on. Kindly help.



what i am stuck at is following.



$$I(m,n) =int_0^{pi/2} (cos x)^{m} sin (n x) dx $$
$$I(m,n)= frac {1}{m+n} + frac{m}{m+n}I(m-1,n-1)$$
after deriving this i am not able to proceed further






calculus definite-integrals







share|cite|improve this question











$endgroup$




$$int_0^{pi/2} (cos x)^{2011} sin (2013 x) dx $$



The value is given to be $frac{1}{2012}$. I am while solving using by parts, but one term is remaining like $frac{1}{2}$, then $frac{1}{3},$ and so on. Kindly help.



what i am stuck at is following.



$$I(m,n) =int_0^{pi/2} (cos x)^{m} sin (n x) dx $$
$$I(m,n)= frac {1}{m+n} + frac{m}{m+n}I(m-1,n-1)$$
after deriving this i am not able to proceed further






calculus definite-integrals




calculus definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 10:26







maveric

















asked Dec 15 '18 at 8:42









mavericmaveric

88012




88012








  • 4




    $begingroup$
    why downvoting?
    $endgroup$
    – maveric
    Dec 15 '18 at 8:45






  • 3




    $begingroup$
    In this case, reasons for the down- and closevotes would be helpful. I do not see one.
    $endgroup$
    – Peter
    Dec 15 '18 at 9:20










  • $begingroup$
    What's meant by stuck? You can not derive that or don't know how to utilize that
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 9:37










  • $begingroup$
    i have reached till this step. i am able to proceed further to answer which is 1/2011
    $endgroup$
    – maveric
    Dec 15 '18 at 9:38










  • $begingroup$
    This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
    $endgroup$
    – Jean Marie
    Dec 15 '18 at 9:51














  • 4




    $begingroup$
    why downvoting?
    $endgroup$
    – maveric
    Dec 15 '18 at 8:45






  • 3




    $begingroup$
    In this case, reasons for the down- and closevotes would be helpful. I do not see one.
    $endgroup$
    – Peter
    Dec 15 '18 at 9:20










  • $begingroup$
    What's meant by stuck? You can not derive that or don't know how to utilize that
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 9:37










  • $begingroup$
    i have reached till this step. i am able to proceed further to answer which is 1/2011
    $endgroup$
    – maveric
    Dec 15 '18 at 9:38










  • $begingroup$
    This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
    $endgroup$
    – Jean Marie
    Dec 15 '18 at 9:51








4




4




$begingroup$
why downvoting?
$endgroup$
– maveric
Dec 15 '18 at 8:45




$begingroup$
why downvoting?
$endgroup$
– maveric
Dec 15 '18 at 8:45




3




3




$begingroup$
In this case, reasons for the down- and closevotes would be helpful. I do not see one.
$endgroup$
– Peter
Dec 15 '18 at 9:20




$begingroup$
In this case, reasons for the down- and closevotes would be helpful. I do not see one.
$endgroup$
– Peter
Dec 15 '18 at 9:20












$begingroup$
What's meant by stuck? You can not derive that or don't know how to utilize that
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:37




$begingroup$
What's meant by stuck? You can not derive that or don't know how to utilize that
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:37












$begingroup$
i have reached till this step. i am able to proceed further to answer which is 1/2011
$endgroup$
– maveric
Dec 15 '18 at 9:38




$begingroup$
i have reached till this step. i am able to proceed further to answer which is 1/2011
$endgroup$
– maveric
Dec 15 '18 at 9:38












$begingroup$
This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
$endgroup$
– Jean Marie
Dec 15 '18 at 9:51




$begingroup$
This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
$endgroup$
– Jean Marie
Dec 15 '18 at 9:51










2 Answers
2






active

oldest

votes


















4












$begingroup$

Hint:



$n=m+2implies$ $$I_{(m,m+2)}=dfrac1{2m+2}+dfrac m{2m+2}I_{(m-1,m+1)}$$



$I_{(0,2)}=1$



$m=1implies$ $$I_{(1,3)}=dfrac14+dfrac14I_{(0,2)}=dfrac12$$



$m=2implies$ $$I_{(2,4)}=dfrac16+dfrac13I_{(1,3)}=dfrac13$$



Use induction for $I_{(m-1,m+1)}=dfrac2{m-1+m+1}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
    $endgroup$
    – maveric
    Dec 15 '18 at 10:13










  • $begingroup$
    @maveric, Yes because of the difference in powers.
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 10:16










  • $begingroup$
    could you ellaborate on induction ?
    $endgroup$
    – maveric
    Dec 15 '18 at 10:27






  • 1




    $begingroup$
    @If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 10:30



















2












$begingroup$

Note that
$$
begin{align}
I(m,n)
&=int_0^{pi/2}cos^m(x)sin(nx),mathrm{d}xtag1\
&=int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}x+int_0^{pi/2}cos^m(x)cos((n-1)x)sin(x),mathrm{d}xtag2\
&=I(m+1,n-1)-frac1{m+1}int_0^{pi/2}cos((n-1)x),mathrm{d}cos^{m+1}(x)tag3\
&=I(m+1,n-1)+frac1{m+1}-frac{n-1}{m+1}int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}xtag4\
&=frac{m-n+2}{m+1},I(m+1,n-1)+frac1{m+1}tag5
end{align}
$$
Explanation:
$(1)$: define $I(m,n)$
$(2)$: $sin(nx)=sin((n-1)x)cos(x)+cos((n-1)x)sin(x)$
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: simplify



Plug in $m=2011$ and $n=2013$ and see what happens.



However, this gives the answer $frac1{2012}$, which is confirmed by Mathematica.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Hint:



    $n=m+2implies$ $$I_{(m,m+2)}=dfrac1{2m+2}+dfrac m{2m+2}I_{(m-1,m+1)}$$



    $I_{(0,2)}=1$



    $m=1implies$ $$I_{(1,3)}=dfrac14+dfrac14I_{(0,2)}=dfrac12$$



    $m=2implies$ $$I_{(2,4)}=dfrac16+dfrac13I_{(1,3)}=dfrac13$$



    Use induction for $I_{(m-1,m+1)}=dfrac2{m-1+m+1}$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
      $endgroup$
      – maveric
      Dec 15 '18 at 10:13










    • $begingroup$
      @maveric, Yes because of the difference in powers.
      $endgroup$
      – lab bhattacharjee
      Dec 15 '18 at 10:16










    • $begingroup$
      could you ellaborate on induction ?
      $endgroup$
      – maveric
      Dec 15 '18 at 10:27






    • 1




      $begingroup$
      @If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
      $endgroup$
      – lab bhattacharjee
      Dec 15 '18 at 10:30
















    4












    $begingroup$

    Hint:



    $n=m+2implies$ $$I_{(m,m+2)}=dfrac1{2m+2}+dfrac m{2m+2}I_{(m-1,m+1)}$$



    $I_{(0,2)}=1$



    $m=1implies$ $$I_{(1,3)}=dfrac14+dfrac14I_{(0,2)}=dfrac12$$



    $m=2implies$ $$I_{(2,4)}=dfrac16+dfrac13I_{(1,3)}=dfrac13$$



    Use induction for $I_{(m-1,m+1)}=dfrac2{m-1+m+1}$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
      $endgroup$
      – maveric
      Dec 15 '18 at 10:13










    • $begingroup$
      @maveric, Yes because of the difference in powers.
      $endgroup$
      – lab bhattacharjee
      Dec 15 '18 at 10:16










    • $begingroup$
      could you ellaborate on induction ?
      $endgroup$
      – maveric
      Dec 15 '18 at 10:27






    • 1




      $begingroup$
      @If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
      $endgroup$
      – lab bhattacharjee
      Dec 15 '18 at 10:30














    4












    4








    4





    $begingroup$

    Hint:



    $n=m+2implies$ $$I_{(m,m+2)}=dfrac1{2m+2}+dfrac m{2m+2}I_{(m-1,m+1)}$$



    $I_{(0,2)}=1$



    $m=1implies$ $$I_{(1,3)}=dfrac14+dfrac14I_{(0,2)}=dfrac12$$



    $m=2implies$ $$I_{(2,4)}=dfrac16+dfrac13I_{(1,3)}=dfrac13$$



    Use induction for $I_{(m-1,m+1)}=dfrac2{m-1+m+1}$






    share|cite|improve this answer









    $endgroup$



    Hint:



    $n=m+2implies$ $$I_{(m,m+2)}=dfrac1{2m+2}+dfrac m{2m+2}I_{(m-1,m+1)}$$



    $I_{(0,2)}=1$



    $m=1implies$ $$I_{(1,3)}=dfrac14+dfrac14I_{(0,2)}=dfrac12$$



    $m=2implies$ $$I_{(2,4)}=dfrac16+dfrac13I_{(1,3)}=dfrac13$$



    Use induction for $I_{(m-1,m+1)}=dfrac2{m-1+m+1}$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 15 '18 at 9:49









    lab bhattacharjeelab bhattacharjee

    226k15157275




    226k15157275












    • $begingroup$
      why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
      $endgroup$
      – maveric
      Dec 15 '18 at 10:13










    • $begingroup$
      @maveric, Yes because of the difference in powers.
      $endgroup$
      – lab bhattacharjee
      Dec 15 '18 at 10:16










    • $begingroup$
      could you ellaborate on induction ?
      $endgroup$
      – maveric
      Dec 15 '18 at 10:27






    • 1




      $begingroup$
      @If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
      $endgroup$
      – lab bhattacharjee
      Dec 15 '18 at 10:30


















    • $begingroup$
      why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
      $endgroup$
      – maveric
      Dec 15 '18 at 10:13










    • $begingroup$
      @maveric, Yes because of the difference in powers.
      $endgroup$
      – lab bhattacharjee
      Dec 15 '18 at 10:16










    • $begingroup$
      could you ellaborate on induction ?
      $endgroup$
      – maveric
      Dec 15 '18 at 10:27






    • 1




      $begingroup$
      @If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
      $endgroup$
      – lab bhattacharjee
      Dec 15 '18 at 10:30
















    $begingroup$
    why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
    $endgroup$
    – maveric
    Dec 15 '18 at 10:13




    $begingroup$
    why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
    $endgroup$
    – maveric
    Dec 15 '18 at 10:13












    $begingroup$
    @maveric, Yes because of the difference in powers.
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 10:16




    $begingroup$
    @maveric, Yes because of the difference in powers.
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 10:16












    $begingroup$
    could you ellaborate on induction ?
    $endgroup$
    – maveric
    Dec 15 '18 at 10:27




    $begingroup$
    could you ellaborate on induction ?
    $endgroup$
    – maveric
    Dec 15 '18 at 10:27




    1




    1




    $begingroup$
    @If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 10:30




    $begingroup$
    @If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 10:30











    2












    $begingroup$

    Note that
    $$
    begin{align}
    I(m,n)
    &=int_0^{pi/2}cos^m(x)sin(nx),mathrm{d}xtag1\
    &=int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}x+int_0^{pi/2}cos^m(x)cos((n-1)x)sin(x),mathrm{d}xtag2\
    &=I(m+1,n-1)-frac1{m+1}int_0^{pi/2}cos((n-1)x),mathrm{d}cos^{m+1}(x)tag3\
    &=I(m+1,n-1)+frac1{m+1}-frac{n-1}{m+1}int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}xtag4\
    &=frac{m-n+2}{m+1},I(m+1,n-1)+frac1{m+1}tag5
    end{align}
    $$
    Explanation:
    $(1)$: define $I(m,n)$
    $(2)$: $sin(nx)=sin((n-1)x)cos(x)+cos((n-1)x)sin(x)$
    $(3)$: prepare to integrate by parts
    $(4)$: integrate by parts
    $(5)$: simplify



    Plug in $m=2011$ and $n=2013$ and see what happens.



    However, this gives the answer $frac1{2012}$, which is confirmed by Mathematica.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Note that
      $$
      begin{align}
      I(m,n)
      &=int_0^{pi/2}cos^m(x)sin(nx),mathrm{d}xtag1\
      &=int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}x+int_0^{pi/2}cos^m(x)cos((n-1)x)sin(x),mathrm{d}xtag2\
      &=I(m+1,n-1)-frac1{m+1}int_0^{pi/2}cos((n-1)x),mathrm{d}cos^{m+1}(x)tag3\
      &=I(m+1,n-1)+frac1{m+1}-frac{n-1}{m+1}int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}xtag4\
      &=frac{m-n+2}{m+1},I(m+1,n-1)+frac1{m+1}tag5
      end{align}
      $$
      Explanation:
      $(1)$: define $I(m,n)$
      $(2)$: $sin(nx)=sin((n-1)x)cos(x)+cos((n-1)x)sin(x)$
      $(3)$: prepare to integrate by parts
      $(4)$: integrate by parts
      $(5)$: simplify



      Plug in $m=2011$ and $n=2013$ and see what happens.



      However, this gives the answer $frac1{2012}$, which is confirmed by Mathematica.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Note that
        $$
        begin{align}
        I(m,n)
        &=int_0^{pi/2}cos^m(x)sin(nx),mathrm{d}xtag1\
        &=int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}x+int_0^{pi/2}cos^m(x)cos((n-1)x)sin(x),mathrm{d}xtag2\
        &=I(m+1,n-1)-frac1{m+1}int_0^{pi/2}cos((n-1)x),mathrm{d}cos^{m+1}(x)tag3\
        &=I(m+1,n-1)+frac1{m+1}-frac{n-1}{m+1}int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}xtag4\
        &=frac{m-n+2}{m+1},I(m+1,n-1)+frac1{m+1}tag5
        end{align}
        $$
        Explanation:
        $(1)$: define $I(m,n)$
        $(2)$: $sin(nx)=sin((n-1)x)cos(x)+cos((n-1)x)sin(x)$
        $(3)$: prepare to integrate by parts
        $(4)$: integrate by parts
        $(5)$: simplify



        Plug in $m=2011$ and $n=2013$ and see what happens.



        However, this gives the answer $frac1{2012}$, which is confirmed by Mathematica.






        share|cite|improve this answer











        $endgroup$



        Note that
        $$
        begin{align}
        I(m,n)
        &=int_0^{pi/2}cos^m(x)sin(nx),mathrm{d}xtag1\
        &=int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}x+int_0^{pi/2}cos^m(x)cos((n-1)x)sin(x),mathrm{d}xtag2\
        &=I(m+1,n-1)-frac1{m+1}int_0^{pi/2}cos((n-1)x),mathrm{d}cos^{m+1}(x)tag3\
        &=I(m+1,n-1)+frac1{m+1}-frac{n-1}{m+1}int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}xtag4\
        &=frac{m-n+2}{m+1},I(m+1,n-1)+frac1{m+1}tag5
        end{align}
        $$
        Explanation:
        $(1)$: define $I(m,n)$
        $(2)$: $sin(nx)=sin((n-1)x)cos(x)+cos((n-1)x)sin(x)$
        $(3)$: prepare to integrate by parts
        $(4)$: integrate by parts
        $(5)$: simplify



        Plug in $m=2011$ and $n=2013$ and see what happens.



        However, this gives the answer $frac1{2012}$, which is confirmed by Mathematica.







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        edited Dec 15 '18 at 10:21

























        answered Dec 15 '18 at 10:13









        robjohn♦robjohn

        268k27309634




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