What is $int_0^{pi/2}sin^7(theta)cos^5(theta)dtheta$












4












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I have to integrate the following:



$int_0^limitsfrac{pi}{2}sin^7(theta)cos^5(theta)dtheta$



I decided to use a $u$ substitution of $u=sin^2(theta)$, and $frac{du}{2}=sin(theta)cos(theta)$



and arrived at this integral



$int_limits{0}^{1}u^3(1-u)^2du$



From here I decided to use integration by parts using $g=u^3$ and $dv=(1-u)^2du$



I get the following:



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-int_limits{0}^{1}(u^2*(1-u)^3)du$$



Repeated again $g=u^2$, and $dv=(1-u)^3du$



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}int_limits{0}^{1}u(1-u)^4$$



Repeating again $g=u$, and $dv=(1-u)^4$



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}biggl[frac{u(1-u)^5}{5}biggr]_0^1-frac{1}5int_limits{0}^{1}(1-u)^5$$



and I get



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}biggl[frac{u(1-u)^5}{5}biggr]_0^1-frac{1}{30}biggl[(1-u)^6biggr]_0^1$$










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  • $begingroup$
    If you just expand $(1-u)^2=u^2+1-2u$, you will be able to solve it more directly.
    $endgroup$
    – Shubham Johri
    Jan 6 at 18:09










  • $begingroup$
    Possible duplicate Integral $int_0^frac{pi}{2} sin^7x cos^5x, dx$
    $endgroup$
    – A.Γ.
    Jan 6 at 18:15












  • $begingroup$
    $$frac 12 B(4,3)$$?
    $endgroup$
    – Digamma
    Jan 7 at 2:45
















4












$begingroup$


I have to integrate the following:



$int_0^limitsfrac{pi}{2}sin^7(theta)cos^5(theta)dtheta$



I decided to use a $u$ substitution of $u=sin^2(theta)$, and $frac{du}{2}=sin(theta)cos(theta)$



and arrived at this integral



$int_limits{0}^{1}u^3(1-u)^2du$



From here I decided to use integration by parts using $g=u^3$ and $dv=(1-u)^2du$



I get the following:



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-int_limits{0}^{1}(u^2*(1-u)^3)du$$



Repeated again $g=u^2$, and $dv=(1-u)^3du$



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}int_limits{0}^{1}u(1-u)^4$$



Repeating again $g=u$, and $dv=(1-u)^4$



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}biggl[frac{u(1-u)^5}{5}biggr]_0^1-frac{1}5int_limits{0}^{1}(1-u)^5$$



and I get



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}biggl[frac{u(1-u)^5}{5}biggr]_0^1-frac{1}{30}biggl[(1-u)^6biggr]_0^1$$










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  • $begingroup$
    If you just expand $(1-u)^2=u^2+1-2u$, you will be able to solve it more directly.
    $endgroup$
    – Shubham Johri
    Jan 6 at 18:09










  • $begingroup$
    Possible duplicate Integral $int_0^frac{pi}{2} sin^7x cos^5x, dx$
    $endgroup$
    – A.Γ.
    Jan 6 at 18:15












  • $begingroup$
    $$frac 12 B(4,3)$$?
    $endgroup$
    – Digamma
    Jan 7 at 2:45














4












4








4





$begingroup$


I have to integrate the following:



$int_0^limitsfrac{pi}{2}sin^7(theta)cos^5(theta)dtheta$



I decided to use a $u$ substitution of $u=sin^2(theta)$, and $frac{du}{2}=sin(theta)cos(theta)$



and arrived at this integral



$int_limits{0}^{1}u^3(1-u)^2du$



From here I decided to use integration by parts using $g=u^3$ and $dv=(1-u)^2du$



I get the following:



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-int_limits{0}^{1}(u^2*(1-u)^3)du$$



Repeated again $g=u^2$, and $dv=(1-u)^3du$



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}int_limits{0}^{1}u(1-u)^4$$



Repeating again $g=u$, and $dv=(1-u)^4$



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}biggl[frac{u(1-u)^5}{5}biggr]_0^1-frac{1}5int_limits{0}^{1}(1-u)^5$$



and I get



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}biggl[frac{u(1-u)^5}{5}biggr]_0^1-frac{1}{30}biggl[(1-u)^6biggr]_0^1$$










share|cite|improve this question











$endgroup$




I have to integrate the following:



$int_0^limitsfrac{pi}{2}sin^7(theta)cos^5(theta)dtheta$



I decided to use a $u$ substitution of $u=sin^2(theta)$, and $frac{du}{2}=sin(theta)cos(theta)$



and arrived at this integral



$int_limits{0}^{1}u^3(1-u)^2du$



From here I decided to use integration by parts using $g=u^3$ and $dv=(1-u)^2du$



I get the following:



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-int_limits{0}^{1}(u^2*(1-u)^3)du$$



Repeated again $g=u^2$, and $dv=(1-u)^3du$



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}int_limits{0}^{1}u(1-u)^4$$



Repeating again $g=u$, and $dv=(1-u)^4$



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}biggl[frac{u(1-u)^5}{5}biggr]_0^1-frac{1}5int_limits{0}^{1}(1-u)^5$$



and I get



$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}biggl[frac{u(1-u)^5}{5}biggr]_0^1-frac{1}{30}biggl[(1-u)^6biggr]_0^1$$







calculus integration trigonometry definite-integrals






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edited Jan 6 at 18:10









JimmyK4542

40.7k245105




40.7k245105










asked Jan 6 at 17:55









EnlightenedFunkyEnlightenedFunky

7471822




7471822












  • $begingroup$
    If you just expand $(1-u)^2=u^2+1-2u$, you will be able to solve it more directly.
    $endgroup$
    – Shubham Johri
    Jan 6 at 18:09










  • $begingroup$
    Possible duplicate Integral $int_0^frac{pi}{2} sin^7x cos^5x, dx$
    $endgroup$
    – A.Γ.
    Jan 6 at 18:15












  • $begingroup$
    $$frac 12 B(4,3)$$?
    $endgroup$
    – Digamma
    Jan 7 at 2:45


















  • $begingroup$
    If you just expand $(1-u)^2=u^2+1-2u$, you will be able to solve it more directly.
    $endgroup$
    – Shubham Johri
    Jan 6 at 18:09










  • $begingroup$
    Possible duplicate Integral $int_0^frac{pi}{2} sin^7x cos^5x, dx$
    $endgroup$
    – A.Γ.
    Jan 6 at 18:15












  • $begingroup$
    $$frac 12 B(4,3)$$?
    $endgroup$
    – Digamma
    Jan 7 at 2:45
















$begingroup$
If you just expand $(1-u)^2=u^2+1-2u$, you will be able to solve it more directly.
$endgroup$
– Shubham Johri
Jan 6 at 18:09




$begingroup$
If you just expand $(1-u)^2=u^2+1-2u$, you will be able to solve it more directly.
$endgroup$
– Shubham Johri
Jan 6 at 18:09












$begingroup$
Possible duplicate Integral $int_0^frac{pi}{2} sin^7x cos^5x, dx$
$endgroup$
– A.Γ.
Jan 6 at 18:15






$begingroup$
Possible duplicate Integral $int_0^frac{pi}{2} sin^7x cos^5x, dx$
$endgroup$
– A.Γ.
Jan 6 at 18:15














$begingroup$
$$frac 12 B(4,3)$$?
$endgroup$
– Digamma
Jan 7 at 2:45




$begingroup$
$$frac 12 B(4,3)$$?
$endgroup$
– Digamma
Jan 7 at 2:45










5 Answers
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9












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I think you complicated the last part, after all you are integrating a polynomial.



$displaystyle int_0^1 u^3(1-u)^2mathop{du}=int_0^1 (u^3-2u^4+u^5)mathop{du}=left[frac{u^4}4-2frac{u^5}5+frac{u^6}6right]_0^1=frac 14-frac 25+frac 16=frac 1{60}$



Also you dropped the coeff $dfrac 12$ from $dfrac{du}2$, the result should be $dfrac 1{120}$






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    5












    $begingroup$

    Note that:
    $$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(theta)sin^{2n+1}(theta)dtheta=frac{m!n!}{(m+n+1)!}$$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I think this is overkill but nevertheless, it's a solution too!
      $endgroup$
      – Frank W.
      Jan 6 at 21:23



















    3












    $begingroup$

    I would just do $u=sintheta$ and $mathrm du=costheta,mathrm dtheta$. Sobegin{align}int_0^{fracpi2}sin^7(theta)cos^5(theta),mathrm dtheta&=int_0^{fracpi2}sin^7(theta)bigl(1-sin^2(theta)bigr)^2cos(theta),mathrm dtheta\&=int_0^1u^7(1-u^2)^2,mathrm du.end{align}I think that it's simpler.






    share|cite|improve this answer











    $endgroup$





















      3












      $begingroup$

      To lower the exponents a bit, notice that substitution $theta mapsto fracpi2-theta$ yields
      $$int_0^{fracpi2} cos^7thetasin^5theta,dtheta = int_0^{fracpi2} sin^7thetacos^5theta,dtheta$$



      so we have
      $$2I = int_0^{fracpi2}sin^5thetacos^5theta(cos^2theta+sin^2theta),dtheta = int_0^{fracpi2}sin^5thetacos^5theta,dtheta = int_0^{fracpi2}sin^5theta(1-sin^2theta)^2costheta,dtheta$$
      Now setting $u = sintheta$ yields
      $$I = frac12 int_0^1u^5(1-u^2)^2,du = frac1{120}$$






      share|cite|improve this answer









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        2












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        As has been noted, this integral can be related to the Beta Function. Here's how.



        Consider the integral
        $$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx$$
        If we make the substitution $t=sin(x)^2$, we have that
        $$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
        $$I(a,b)=frac12int_0^1t^{frac{a+1}2-1}(1-t)^{frac{b+1}2-1}mathrm dt$$
        We then recall the definition of the Beta function
        $$mathrm B(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm dt=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
        Where $Gamma(s)$ is the Gamma function. Technically it is defined by
        $$Gamma(s)=int_0^infty x^{s-1}e^{-x}mathrm dx,qquad mathrm{Re}(s)>0$$
        But in the case that $s$ is a (positive) integer,
        $$Gamma(s)=(s-1)!$$
        So without further adieu,
        $$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
        We then see that your integral is
        $$I(7,5)=frac{Gamma(4)Gamma(3)}{2Gamma(7)}$$
        $$I(7,5)=frac1{120}$$






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          5 Answers
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          5 Answers
          5






          active

          oldest

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          active

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          active

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          9












          $begingroup$

          I think you complicated the last part, after all you are integrating a polynomial.



          $displaystyle int_0^1 u^3(1-u)^2mathop{du}=int_0^1 (u^3-2u^4+u^5)mathop{du}=left[frac{u^4}4-2frac{u^5}5+frac{u^6}6right]_0^1=frac 14-frac 25+frac 16=frac 1{60}$



          Also you dropped the coeff $dfrac 12$ from $dfrac{du}2$, the result should be $dfrac 1{120}$






          share|cite|improve this answer









          $endgroup$


















            9












            $begingroup$

            I think you complicated the last part, after all you are integrating a polynomial.



            $displaystyle int_0^1 u^3(1-u)^2mathop{du}=int_0^1 (u^3-2u^4+u^5)mathop{du}=left[frac{u^4}4-2frac{u^5}5+frac{u^6}6right]_0^1=frac 14-frac 25+frac 16=frac 1{60}$



            Also you dropped the coeff $dfrac 12$ from $dfrac{du}2$, the result should be $dfrac 1{120}$






            share|cite|improve this answer









            $endgroup$
















              9












              9








              9





              $begingroup$

              I think you complicated the last part, after all you are integrating a polynomial.



              $displaystyle int_0^1 u^3(1-u)^2mathop{du}=int_0^1 (u^3-2u^4+u^5)mathop{du}=left[frac{u^4}4-2frac{u^5}5+frac{u^6}6right]_0^1=frac 14-frac 25+frac 16=frac 1{60}$



              Also you dropped the coeff $dfrac 12$ from $dfrac{du}2$, the result should be $dfrac 1{120}$






              share|cite|improve this answer









              $endgroup$



              I think you complicated the last part, after all you are integrating a polynomial.



              $displaystyle int_0^1 u^3(1-u)^2mathop{du}=int_0^1 (u^3-2u^4+u^5)mathop{du}=left[frac{u^4}4-2frac{u^5}5+frac{u^6}6right]_0^1=frac 14-frac 25+frac 16=frac 1{60}$



              Also you dropped the coeff $dfrac 12$ from $dfrac{du}2$, the result should be $dfrac 1{120}$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 6 at 18:10









              zwimzwim

              11.7k729




              11.7k729























                  5












                  $begingroup$

                  Note that:
                  $$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(theta)sin^{2n+1}(theta)dtheta=frac{m!n!}{(m+n+1)!}$$






                  share|cite|improve this answer









                  $endgroup$









                  • 1




                    $begingroup$
                    I think this is overkill but nevertheless, it's a solution too!
                    $endgroup$
                    – Frank W.
                    Jan 6 at 21:23
















                  5












                  $begingroup$

                  Note that:
                  $$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(theta)sin^{2n+1}(theta)dtheta=frac{m!n!}{(m+n+1)!}$$






                  share|cite|improve this answer









                  $endgroup$









                  • 1




                    $begingroup$
                    I think this is overkill but nevertheless, it's a solution too!
                    $endgroup$
                    – Frank W.
                    Jan 6 at 21:23














                  5












                  5








                  5





                  $begingroup$

                  Note that:
                  $$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(theta)sin^{2n+1}(theta)dtheta=frac{m!n!}{(m+n+1)!}$$






                  share|cite|improve this answer









                  $endgroup$



                  Note that:
                  $$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(theta)sin^{2n+1}(theta)dtheta=frac{m!n!}{(m+n+1)!}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 18:29









                  Henry LeeHenry Lee

                  1,855219




                  1,855219








                  • 1




                    $begingroup$
                    I think this is overkill but nevertheless, it's a solution too!
                    $endgroup$
                    – Frank W.
                    Jan 6 at 21:23














                  • 1




                    $begingroup$
                    I think this is overkill but nevertheless, it's a solution too!
                    $endgroup$
                    – Frank W.
                    Jan 6 at 21:23








                  1




                  1




                  $begingroup$
                  I think this is overkill but nevertheless, it's a solution too!
                  $endgroup$
                  – Frank W.
                  Jan 6 at 21:23




                  $begingroup$
                  I think this is overkill but nevertheless, it's a solution too!
                  $endgroup$
                  – Frank W.
                  Jan 6 at 21:23











                  3












                  $begingroup$

                  I would just do $u=sintheta$ and $mathrm du=costheta,mathrm dtheta$. Sobegin{align}int_0^{fracpi2}sin^7(theta)cos^5(theta),mathrm dtheta&=int_0^{fracpi2}sin^7(theta)bigl(1-sin^2(theta)bigr)^2cos(theta),mathrm dtheta\&=int_0^1u^7(1-u^2)^2,mathrm du.end{align}I think that it's simpler.






                  share|cite|improve this answer











                  $endgroup$


















                    3












                    $begingroup$

                    I would just do $u=sintheta$ and $mathrm du=costheta,mathrm dtheta$. Sobegin{align}int_0^{fracpi2}sin^7(theta)cos^5(theta),mathrm dtheta&=int_0^{fracpi2}sin^7(theta)bigl(1-sin^2(theta)bigr)^2cos(theta),mathrm dtheta\&=int_0^1u^7(1-u^2)^2,mathrm du.end{align}I think that it's simpler.






                    share|cite|improve this answer











                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      I would just do $u=sintheta$ and $mathrm du=costheta,mathrm dtheta$. Sobegin{align}int_0^{fracpi2}sin^7(theta)cos^5(theta),mathrm dtheta&=int_0^{fracpi2}sin^7(theta)bigl(1-sin^2(theta)bigr)^2cos(theta),mathrm dtheta\&=int_0^1u^7(1-u^2)^2,mathrm du.end{align}I think that it's simpler.






                      share|cite|improve this answer











                      $endgroup$



                      I would just do $u=sintheta$ and $mathrm du=costheta,mathrm dtheta$. Sobegin{align}int_0^{fracpi2}sin^7(theta)cos^5(theta),mathrm dtheta&=int_0^{fracpi2}sin^7(theta)bigl(1-sin^2(theta)bigr)^2cos(theta),mathrm dtheta\&=int_0^1u^7(1-u^2)^2,mathrm du.end{align}I think that it's simpler.







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                      share|cite|improve this answer








                      edited Jan 6 at 21:24

























                      answered Jan 6 at 18:03









                      José Carlos SantosJosé Carlos Santos

                      154k22124227




                      154k22124227























                          3












                          $begingroup$

                          To lower the exponents a bit, notice that substitution $theta mapsto fracpi2-theta$ yields
                          $$int_0^{fracpi2} cos^7thetasin^5theta,dtheta = int_0^{fracpi2} sin^7thetacos^5theta,dtheta$$



                          so we have
                          $$2I = int_0^{fracpi2}sin^5thetacos^5theta(cos^2theta+sin^2theta),dtheta = int_0^{fracpi2}sin^5thetacos^5theta,dtheta = int_0^{fracpi2}sin^5theta(1-sin^2theta)^2costheta,dtheta$$
                          Now setting $u = sintheta$ yields
                          $$I = frac12 int_0^1u^5(1-u^2)^2,du = frac1{120}$$






                          share|cite|improve this answer









                          $endgroup$


















                            3












                            $begingroup$

                            To lower the exponents a bit, notice that substitution $theta mapsto fracpi2-theta$ yields
                            $$int_0^{fracpi2} cos^7thetasin^5theta,dtheta = int_0^{fracpi2} sin^7thetacos^5theta,dtheta$$



                            so we have
                            $$2I = int_0^{fracpi2}sin^5thetacos^5theta(cos^2theta+sin^2theta),dtheta = int_0^{fracpi2}sin^5thetacos^5theta,dtheta = int_0^{fracpi2}sin^5theta(1-sin^2theta)^2costheta,dtheta$$
                            Now setting $u = sintheta$ yields
                            $$I = frac12 int_0^1u^5(1-u^2)^2,du = frac1{120}$$






                            share|cite|improve this answer









                            $endgroup$
















                              3












                              3








                              3





                              $begingroup$

                              To lower the exponents a bit, notice that substitution $theta mapsto fracpi2-theta$ yields
                              $$int_0^{fracpi2} cos^7thetasin^5theta,dtheta = int_0^{fracpi2} sin^7thetacos^5theta,dtheta$$



                              so we have
                              $$2I = int_0^{fracpi2}sin^5thetacos^5theta(cos^2theta+sin^2theta),dtheta = int_0^{fracpi2}sin^5thetacos^5theta,dtheta = int_0^{fracpi2}sin^5theta(1-sin^2theta)^2costheta,dtheta$$
                              Now setting $u = sintheta$ yields
                              $$I = frac12 int_0^1u^5(1-u^2)^2,du = frac1{120}$$






                              share|cite|improve this answer









                              $endgroup$



                              To lower the exponents a bit, notice that substitution $theta mapsto fracpi2-theta$ yields
                              $$int_0^{fracpi2} cos^7thetasin^5theta,dtheta = int_0^{fracpi2} sin^7thetacos^5theta,dtheta$$



                              so we have
                              $$2I = int_0^{fracpi2}sin^5thetacos^5theta(cos^2theta+sin^2theta),dtheta = int_0^{fracpi2}sin^5thetacos^5theta,dtheta = int_0^{fracpi2}sin^5theta(1-sin^2theta)^2costheta,dtheta$$
                              Now setting $u = sintheta$ yields
                              $$I = frac12 int_0^1u^5(1-u^2)^2,du = frac1{120}$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 6 at 22:07









                              mechanodroidmechanodroid

                              27.1k62446




                              27.1k62446























                                  2












                                  $begingroup$

                                  As has been noted, this integral can be related to the Beta Function. Here's how.



                                  Consider the integral
                                  $$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx$$
                                  If we make the substitution $t=sin(x)^2$, we have that
                                  $$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
                                  $$I(a,b)=frac12int_0^1t^{frac{a+1}2-1}(1-t)^{frac{b+1}2-1}mathrm dt$$
                                  We then recall the definition of the Beta function
                                  $$mathrm B(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm dt=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
                                  Where $Gamma(s)$ is the Gamma function. Technically it is defined by
                                  $$Gamma(s)=int_0^infty x^{s-1}e^{-x}mathrm dx,qquad mathrm{Re}(s)>0$$
                                  But in the case that $s$ is a (positive) integer,
                                  $$Gamma(s)=(s-1)!$$
                                  So without further adieu,
                                  $$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
                                  We then see that your integral is
                                  $$I(7,5)=frac{Gamma(4)Gamma(3)}{2Gamma(7)}$$
                                  $$I(7,5)=frac1{120}$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    2












                                    $begingroup$

                                    As has been noted, this integral can be related to the Beta Function. Here's how.



                                    Consider the integral
                                    $$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx$$
                                    If we make the substitution $t=sin(x)^2$, we have that
                                    $$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
                                    $$I(a,b)=frac12int_0^1t^{frac{a+1}2-1}(1-t)^{frac{b+1}2-1}mathrm dt$$
                                    We then recall the definition of the Beta function
                                    $$mathrm B(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm dt=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
                                    Where $Gamma(s)$ is the Gamma function. Technically it is defined by
                                    $$Gamma(s)=int_0^infty x^{s-1}e^{-x}mathrm dx,qquad mathrm{Re}(s)>0$$
                                    But in the case that $s$ is a (positive) integer,
                                    $$Gamma(s)=(s-1)!$$
                                    So without further adieu,
                                    $$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
                                    We then see that your integral is
                                    $$I(7,5)=frac{Gamma(4)Gamma(3)}{2Gamma(7)}$$
                                    $$I(7,5)=frac1{120}$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      As has been noted, this integral can be related to the Beta Function. Here's how.



                                      Consider the integral
                                      $$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx$$
                                      If we make the substitution $t=sin(x)^2$, we have that
                                      $$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
                                      $$I(a,b)=frac12int_0^1t^{frac{a+1}2-1}(1-t)^{frac{b+1}2-1}mathrm dt$$
                                      We then recall the definition of the Beta function
                                      $$mathrm B(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm dt=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
                                      Where $Gamma(s)$ is the Gamma function. Technically it is defined by
                                      $$Gamma(s)=int_0^infty x^{s-1}e^{-x}mathrm dx,qquad mathrm{Re}(s)>0$$
                                      But in the case that $s$ is a (positive) integer,
                                      $$Gamma(s)=(s-1)!$$
                                      So without further adieu,
                                      $$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
                                      We then see that your integral is
                                      $$I(7,5)=frac{Gamma(4)Gamma(3)}{2Gamma(7)}$$
                                      $$I(7,5)=frac1{120}$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      As has been noted, this integral can be related to the Beta Function. Here's how.



                                      Consider the integral
                                      $$I(a,b)=int_0^{pi/2}sin(x)^acos(x)^bmathrm dx$$
                                      If we make the substitution $t=sin(x)^2$, we have that
                                      $$I(a,b)=frac12int_0^1t^{frac{a-1}2}(1-t)^{frac{b-1}2}mathrm dt$$
                                      $$I(a,b)=frac12int_0^1t^{frac{a+1}2-1}(1-t)^{frac{b+1}2-1}mathrm dt$$
                                      We then recall the definition of the Beta function
                                      $$mathrm B(x,y)=int_0^1t^{x-1}(1-t)^{y-1}mathrm dt=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
                                      Where $Gamma(s)$ is the Gamma function. Technically it is defined by
                                      $$Gamma(s)=int_0^infty x^{s-1}e^{-x}mathrm dx,qquad mathrm{Re}(s)>0$$
                                      But in the case that $s$ is a (positive) integer,
                                      $$Gamma(s)=(s-1)!$$
                                      So without further adieu,
                                      $$I(a,b)=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
                                      We then see that your integral is
                                      $$I(7,5)=frac{Gamma(4)Gamma(3)}{2Gamma(7)}$$
                                      $$I(7,5)=frac1{120}$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 6 at 23:03









                                      clathratusclathratus

                                      3,646332




                                      3,646332






























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