Mixing
What is $int_0^{pi/2}sin^7(theta)cos^5(theta)dtheta$
$begingroup$
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
$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}biggl[frac{u(1-u)^5}{5}biggr]_0^1-frac{1}{30}biggl[(1-u)^6biggr]_0^1$$
calculus integration trigonometry definite-integrals
edited Jan 6 at 18:10
JimmyK4542
40.7k245105
asked Jan 6 at 17:55
EnlightenedFunkyEnlightenedFunky
7471822
$endgroup$
add a comment |
$begingroup$
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
$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}biggl[frac{u(1-u)^5}{5}biggr]_0^1-frac{1}{30}biggl[(1-u)^6biggr]_0^1$$
calculus integration trigonometry definite-integrals
edited Jan 6 at 18:10
JimmyK4542
40.7k245105
asked Jan 6 at 17:55
EnlightenedFunkyEnlightenedFunky
7471822
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$biggl[frac{u^3*(1-u)^3}{3}biggr]_0^1-biggl[frac{u^2*(1-u)^4}{4}biggr]_0^1+frac{1}{2}biggl[frac{u(1-u)^5}{5}biggr]_0^1-frac{1}{30}biggl[(1-u)^6biggr]_0^1$$
calculus integration trigonometry definite-integrals
edited Jan 6 at 18:10
JimmyK4542
40.7k245105
asked Jan 6 at 17:55
EnlightenedFunkyEnlightenedFunky
7471822
$endgroup$
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
calculus integration trigonometry definite-integrals
edited Jan 6 at 18:10
JimmyK4542
40.7k245105
asked Jan 6 at 17:55
EnlightenedFunkyEnlightenedFunky
7471822
edited Jan 6 at 18:10
JimmyK4542
40.7k245105
asked Jan 6 at 17:55
EnlightenedFunkyEnlightenedFunky
7471822
edited Jan 6 at 18:10
JimmyK4542
40.7k245105
edited Jan 6 at 18:10
JimmyK4542
40.7k245105
edited Jan 6 at 18:10
JimmyK4542
40.7k245105
40.7k245105
asked Jan 6 at 17:55
EnlightenedFunkyEnlightenedFunky
7471822
asked Jan 6 at 17:55
EnlightenedFunkyEnlightenedFunky
7471822
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
add a comment |
$begingroup$
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
add a comment |
5 Answers
5
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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 $dfrac 1{120}$
answered Jan 6 at 18:10
zwimzwim
11.8k730
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add a comment |
$begingroup$
Note that:
$$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(theta)sin^{2n+1}(theta)dtheta=frac{m!n!}{(m+n+1)!}$$
answered Jan 6 at 18:29
Henry LeeHenry Lee
1,884219
$endgroup$
1
$begingroup$
I think this is overkill but nevertheless, it's a solution too!
$endgroup$
– Frank W.
Jan 6 at 21:23
add a comment |
$begingroup$
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.
answered Jan 6 at 18:03
José Carlos SantosJosé Carlos Santos
157k22126227
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add a comment |
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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
$$I = frac12 int_0^1u^5(1-u^2)^2,du = frac1{120}$$
answered Jan 6 at 22:07
mechanodroidmechanodroid
27.3k62446
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add a comment |
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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
$$I(7,5)=frac{Gamma(4)Gamma(3)}{2Gamma(7)}$$
$$I(7,5)=frac1{120}$$
answered Jan 6 at 23:03
clathratusclathratus
3,767333
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $dfrac 1{120}$
answered Jan 6 at 18:10
zwimzwim
11.8k730
$endgroup$
add a comment |
$begingroup$
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 $dfrac 1{120}$
answered Jan 6 at 18:10
zwimzwim
11.8k730
$endgroup$
add a comment |
$begingroup$
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 $dfrac 1{120}$
answered Jan 6 at 18:10
zwimzwim
11.8k730
$endgroup$
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 $dfrac 1{120}$
answered Jan 6 at 18:10
zwimzwim
11.8k730
answered Jan 6 at 18:10
zwimzwim
11.8k730
answered Jan 6 at 18:10
zwimzwim
11.8k730
answered Jan 6 at 18:10
zwimzwim
11.8k730
11.8k730
add a comment |
add a comment |
$begingroup$
Note that:
$$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(theta)sin^{2n+1}(theta)dtheta=frac{m!n!}{(m+n+1)!}$$
answered Jan 6 at 18:29
Henry LeeHenry Lee
1,884219
$endgroup$
1
$begingroup$
I think this is overkill but nevertheless, it's a solution too!
$endgroup$
– Frank W.
Jan 6 at 21:23
add a comment |
$begingroup$
Note that:
$$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(theta)sin^{2n+1}(theta)dtheta=frac{m!n!}{(m+n+1)!}$$
answered Jan 6 at 18:29
Henry LeeHenry Lee
1,884219
$endgroup$
1
$begingroup$
I think this is overkill but nevertheless, it's a solution too!
$endgroup$
– Frank W.
Jan 6 at 21:23
add a comment |
$begingroup$
Note that:
$$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(theta)sin^{2n+1}(theta)dtheta=frac{m!n!}{(m+n+1)!}$$
answered Jan 6 at 18:29
Henry LeeHenry Lee
1,884219
$endgroup$
Note that:
$$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(theta)sin^{2n+1}(theta)dtheta=frac{m!n!}{(m+n+1)!}$$
answered Jan 6 at 18:29
Henry LeeHenry Lee
1,884219
answered Jan 6 at 18:29
Henry LeeHenry Lee
1,884219
answered Jan 6 at 18:29
Henry LeeHenry Lee
1,884219
answered Jan 6 at 18:29
Henry LeeHenry Lee
1,884219
1,884219
1
$begingroup$
I think this is overkill but nevertheless, it's a solution too!
$endgroup$
– Frank W.
Jan 6 at 21:23
add a comment |
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
add a comment |
$begingroup$
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.
answered Jan 6 at 18:03
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 6 at 18:03
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 6 at 18:03
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
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.
answered Jan 6 at 18:03
José Carlos SantosJosé Carlos Santos
157k22126227
edited Jan 6 at 21:24
answered Jan 6 at 18:03
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 6 at 18:03
José Carlos SantosJosé Carlos Santos
157k22126227
answered Jan 6 at 18:03
José Carlos SantosJosé Carlos Santos
157k22126227
157k22126227
add a comment |
add a comment |
$begingroup$
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
$$I = frac12 int_0^1u^5(1-u^2)^2,du = frac1{120}$$
answered Jan 6 at 22:07
mechanodroidmechanodroid
27.3k62446
$endgroup$
add a comment |
$begingroup$
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
$$I = frac12 int_0^1u^5(1-u^2)^2,du = frac1{120}$$
answered Jan 6 at 22:07
mechanodroidmechanodroid
27.3k62446
$endgroup$
add a comment |
$begingroup$
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
$$I = frac12 int_0^1u^5(1-u^2)^2,du = frac1{120}$$
answered Jan 6 at 22:07
mechanodroidmechanodroid
27.3k62446
$endgroup$
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
$$I = frac12 int_0^1u^5(1-u^2)^2,du = frac1{120}$$
answered Jan 6 at 22:07
mechanodroidmechanodroid
27.3k62446
answered Jan 6 at 22:07
mechanodroidmechanodroid
27.3k62446
answered Jan 6 at 22:07
mechanodroidmechanodroid
27.3k62446
answered Jan 6 at 22:07
mechanodroidmechanodroid
27.3k62446
27.3k62446
add a comment |
add a comment |
$begingroup$
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
$$I(7,5)=frac{Gamma(4)Gamma(3)}{2Gamma(7)}$$
$$I(7,5)=frac1{120}$$
answered Jan 6 at 23:03
clathratusclathratus
3,767333
$endgroup$
add a comment |
$begingroup$
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
$$I(7,5)=frac{Gamma(4)Gamma(3)}{2Gamma(7)}$$
$$I(7,5)=frac1{120}$$
answered Jan 6 at 23:03
clathratusclathratus
3,767333
$endgroup$
add a comment |
$begingroup$
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
$$I(7,5)=frac{Gamma(4)Gamma(3)}{2Gamma(7)}$$
$$I(7,5)=frac1{120}$$
answered Jan 6 at 23:03
clathratusclathratus
3,767333
$endgroup$
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
$$I(7,5)=frac{Gamma(4)Gamma(3)}{2Gamma(7)}$$
$$I(7,5)=frac1{120}$$
answered Jan 6 at 23:03
clathratusclathratus
3,767333
answered Jan 6 at 23:03
clathratusclathratus
3,767333
answered Jan 6 at 23:03
clathratusclathratus
3,767333
answered Jan 6 at 23:03
clathratusclathratus
3,767333
3,767333
add a comment |
add a comment |
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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