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How to calculate $int_0^pi/2x^2sqrtcos (x)dx$
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I've tried everything. I made $x = pi / 2 - y$. I looked up the questions here on MSE and did not see anything similar.
I thought I'd use complex variables, but I do not even know where to start.
calculus integration definite-integrals
edited Jul 15 at 22:48
amWhy
189k25219431
asked Jul 15 at 22:44
MathFacts
2,9921425
 |Â
show 2 more comments
up vote
3
down vote
favorite
I've tried everything. I made $x = pi / 2 - y$. I looked up the questions here on MSE and did not see anything similar.
I thought I'd use complex variables, but I do not even know where to start.
calculus integration definite-integrals
edited Jul 15 at 22:48
amWhy
189k25219431
asked Jul 15 at 22:44
MathFacts
2,9921425
1
I don't think that you will be able to find the antiderivative of $x^2sqrtcos(x)$: wolframAlpha gives that "monster"
– Holo
Jul 15 at 22:50
Where is the monster? An result containing the known elementary functions is impossible.
– Dr. Sonnhard Graubner
Jul 15 at 22:52
1
You can use a numerical way to catch this.
– Dr. Sonnhard Graubner
Jul 15 at 22:53
I want a certain value of integral.
– MathFacts
Jul 15 at 22:55
Here it is $$.71728411711845890580$$
– Dr. Sonnhard Graubner
Jul 15 at 23:01
 |Â
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I've tried everything. I made $x = pi / 2 - y$. I looked up the questions here on MSE and did not see anything similar.
I thought I'd use complex variables, but I do not even know where to start.
calculus integration definite-integrals
edited Jul 15 at 22:48
amWhy
189k25219431
asked Jul 15 at 22:44
MathFacts
2,9921425
I've tried everything. I made $x = pi / 2 - y$. I looked up the questions here on MSE and did not see anything similar.
I thought I'd use complex variables, but I do not even know where to start.
calculus integration definite-integrals
edited Jul 15 at 22:48
amWhy
189k25219431
asked Jul 15 at 22:44
MathFacts
2,9921425
edited Jul 15 at 22:48
amWhy
189k25219431
edited Jul 15 at 22:48
amWhy
189k25219431
edited Jul 15 at 22:48
amWhy
189k25219431
189k25219431
asked Jul 15 at 22:44
MathFacts
2,9921425
asked Jul 15 at 22:44
MathFacts
2,9921425
asked Jul 15 at 22:44
MathFacts
2,9921425
2,9921425
1
I don't think that you will be able to find the antiderivative of $x^2sqrtcos(x)$: wolframAlpha gives that "monster"
– Holo
Jul 15 at 22:50
Where is the monster? An result containing the known elementary functions is impossible.
– Dr. Sonnhard Graubner
Jul 15 at 22:52
1
You can use a numerical way to catch this.
– Dr. Sonnhard Graubner
Jul 15 at 22:53
I want a certain value of integral.
– MathFacts
Jul 15 at 22:55
Here it is $$.71728411711845890580$$
– Dr. Sonnhard Graubner
Jul 15 at 23:01
 |Â
show 2 more comments
1
I don't think that you will be able to find the antiderivative of $x^2sqrtcos(x)$: wolframAlpha gives that "monster"
– Holo
Jul 15 at 22:50
Where is the monster? An result containing the known elementary functions is impossible.
– Dr. Sonnhard Graubner
Jul 15 at 22:52
1
You can use a numerical way to catch this.
– Dr. Sonnhard Graubner
Jul 15 at 22:53
I want a certain value of integral.
– MathFacts
Jul 15 at 22:55
Here it is $$.71728411711845890580$$
– Dr. Sonnhard Graubner
Jul 15 at 23:01
1
1
I don't think that you will be able to find the antiderivative of $x^2sqrtcos(x)$: wolframAlpha gives that "monster"
– Holo
Jul 15 at 22:50
I don't think that you will be able to find the antiderivative of $x^2sqrtcos(x)$: wolframAlpha gives that "monster"
– Holo
Jul 15 at 22:50
Where is the monster? An result containing the known elementary functions is impossible.
– Dr. Sonnhard Graubner
Jul 15 at 22:52
Where is the monster? An result containing the known elementary functions is impossible.
– Dr. Sonnhard Graubner
Jul 15 at 22:52
1
1
You can use a numerical way to catch this.
– Dr. Sonnhard Graubner
Jul 15 at 22:53
You can use a numerical way to catch this.
– Dr. Sonnhard Graubner
Jul 15 at 22:53
I want a certain value of integral.
– MathFacts
Jul 15 at 22:55
I want a certain value of integral.
– MathFacts
Jul 15 at 22:55
Here it is $$.71728411711845890580$$
– Dr. Sonnhard Graubner
Jul 15 at 23:01
Here it is $$.71728411711845890580$$
– Dr. Sonnhard Graubner
Jul 15 at 23:01
 |Â
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
12
down vote
In the question I asked here Integral $int_0^fracpi2 x^2 sqrtsin x , dx$ @Frank Wei showed a way to evaluate your integral. I will try to show another method with the approach
I started the integral I posted there. $$I=int_0^fracpi2 x^2 sqrtcosxdx = int_0^fracpi2x^2 (1+tan^2 (x))^-frac14dx$$ substituting $tan x=y, $ we get $$ int_0^infty arctan^2 (x) (1+x^2)^-frac54 dx$$ $$I=-frac14int_0^inftylog^2left(frac1-ix1+ixright)(1+x^2)^-frac54 , dx$$ with $frac1-ix1+ix=y, $ we get the integral to be: $$I=Releft( - fraci 8sqrt 2int_-1^1 log^2 (y) , y^-frac54 sqrt1+y , dyright) $$Now we split the integral from $int_-1^0$ and $int_0^1 $substituting $y=-x$, the second one vanishes being purely imaginary, so we are left with:$$I=Re left(frac1+i16int_0^1 x^-frac54 sqrt1+x (-pi^2 +2 i pi log x +log^2 x) dxright) $$ $$I=-fracpi^216 int_0^1 x^-frac14-1(1-x)^frac32-1dx - fracpi 8int_0^1 x^-frac14-1(1-x)^frac32-1 log x,dx+frac1 16 int_0^1 x^-frac14-1(1-x)^frac32-1log^2(x),dx$$ We can evaluate these integrals using beta function, indeed $$I=-fracpi^2 16 Bleft(-frac14, frac32right) -fracpi 8fracddz Bleft(z, frac32right)big|_z=-frac14+frac1 16fracd^2 dz^2 Bleft(z, frac32right)big|_z=-frac14$$ I let you do the algebra in order to get $$I=sqrt 2pi^frac32fracpi^2+8G-16Gamma^2left(frac14right)$$
edited Jul 20 at 0:09
MathFacts
2,9921425
answered Jul 16 at 0:24
Zacky
2,2701327
1
Where $G$ is Catalan's constant, as per @FrankWei's answer
– peter a g
Jul 16 at 1:11
1
This is pure beauty ! Thanks for providing such an answer. Cheers.
– Claude Leibovici
Jul 16 at 5:19
Thank you for those kind words!
– Zacky
Jul 16 at 5:37
Cannot we use some other methods ? What if I split up $x^2$ as $-xleft(dfracpi2 -x right) + dfracpi x2$ ? I am wondering if that'll lead to a simple solution. Your solution is too advanced for me to understand.
– DarkKnight
Jul 16 at 6:12
It seems that $int_0^fracpi2 xsqrtcosx ,dx$ doesnt have a nice closed form so I doubt that would be easier. Surely there may be other approaches to this integral, have you seen @FrankWei's answer?
– Zacky
Jul 16 at 9:45
 |Â
show 1 more comment
up vote
5
down vote
Just for the fun of it !
Since Dahaka provided a splendid and exact solution, I have been wondering what would give the approximation
$$cos(x) simeqfracpi ^2-4x^2pi ^2+x^2qquad (-frac pi 2 leq xleqfrac pi 2)$$ So, considering
$$I=int_0^frac pi 2x^2 sqrtfracpi ^2-4 x^2x^2+pi ^2,dx$$ the antiderivative expresses in terms of elliptic integrals but, using bounds, we get the simple
$$I=frac pi ^36 left(9 Eleft(-frac14right)-10
Kleft(-frac14right)right)approx 0.71832$$ which is in error by $0.14$%.
answered Jul 16 at 5:18
Claude Leibovici
112k1055126
add a comment |Â
up vote
2
down vote
To find a primitive is arduous with the usual methods.
We could provide a rough approximation with the midpoint formula. Let $f(x)=x^2sqrtcos x$ $$int_0^pi/2f(x)approx fracpi2fbigg(fracpi4bigg)=0.8148.$$
If you want a more accurate approximation, a simple way is through the Legendre polynomials. The Legendre polynomias are given by $$P_n(x)=frac12^nn!fracd^ndx^n(x^2-1)^n.$$ We place $n=2$, $P_2(x)=frac12(3x^2-1)$. The quadrature formula, in this case, is
$$
int_-1^1f(x);dxapproxsum_i=1^2 A_if(x_i),
$$
where $$A_i=int_-1^1fracP_2(x)(x-x_i)P'_2(x_i);dx$$
and $x_i$ for $i=1,2$ are $P_2(x)$ roots.
Now, $A_1=A_2=1$ and $x_1=-frac1sqrt3$, $x_2=frac1sqrt3$.
Therefore
$$
int_a^b f(x);dx=fracb-a2int_-1^1 fbigg(fraca+b2+tfracb-a2bigg);dtapprox fracb-a2bigg[1cdot f(y_1)+1cdot f(y_2)bigg],
$$
where
$$
y_1=fraca+b2-frac1sqrt3fracb-a2quad textandquad y_2=fraca+b2+frac1sqrt3fracb-a2.
$$
Then
$$int_0^pi/2 f(x);dxapprox 0.77$$
answered Jul 15 at 22:58
Jack J.
3661317
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
In the question I asked here Integral $int_0^fracpi2 x^2 sqrtsin x , dx$ @Frank Wei showed a way to evaluate your integral. I will try to show another method with the approach
I started the integral I posted there. $$I=int_0^fracpi2 x^2 sqrtcosxdx = int_0^fracpi2x^2 (1+tan^2 (x))^-frac14dx$$ substituting $tan x=y, $ we get $$ int_0^infty arctan^2 (x) (1+x^2)^-frac54 dx$$ $$I=-frac14int_0^inftylog^2left(frac1-ix1+ixright)(1+x^2)^-frac54 , dx$$ with $frac1-ix1+ix=y, $ we get the integral to be: $$I=Releft( - fraci 8sqrt 2int_-1^1 log^2 (y) , y^-frac54 sqrt1+y , dyright) $$Now we split the integral from $int_-1^0$ and $int_0^1 $substituting $y=-x$, the second one vanishes being purely imaginary, so we are left with:$$I=Re left(frac1+i16int_0^1 x^-frac54 sqrt1+x (-pi^2 +2 i pi log x +log^2 x) dxright) $$ $$I=-fracpi^216 int_0^1 x^-frac14-1(1-x)^frac32-1dx - fracpi 8int_0^1 x^-frac14-1(1-x)^frac32-1 log x,dx+frac1 16 int_0^1 x^-frac14-1(1-x)^frac32-1log^2(x),dx$$ We can evaluate these integrals using beta function, indeed $$I=-fracpi^2 16 Bleft(-frac14, frac32right) -fracpi 8fracddz Bleft(z, frac32right)big|_z=-frac14+frac1 16fracd^2 dz^2 Bleft(z, frac32right)big|_z=-frac14$$ I let you do the algebra in order to get $$I=sqrt 2pi^frac32fracpi^2+8G-16Gamma^2left(frac14right)$$
edited Jul 20 at 0:09
MathFacts
2,9921425
answered Jul 16 at 0:24
Zacky
2,2701327
1
Where $G$ is Catalan's constant, as per @FrankWei's answer
– peter a g
Jul 16 at 1:11
1
This is pure beauty ! Thanks for providing such an answer. Cheers.
– Claude Leibovici
Jul 16 at 5:19
Thank you for those kind words!
– Zacky
Jul 16 at 5:37
Cannot we use some other methods ? What if I split up $x^2$ as $-xleft(dfracpi2 -x right) + dfracpi x2$ ? I am wondering if that'll lead to a simple solution. Your solution is too advanced for me to understand.
– DarkKnight
Jul 16 at 6:12
It seems that $int_0^fracpi2 xsqrtcosx ,dx$ doesnt have a nice closed form so I doubt that would be easier. Surely there may be other approaches to this integral, have you seen @FrankWei's answer?
– Zacky
Jul 16 at 9:45
 |Â
show 1 more comment
up vote
12
down vote
In the question I asked here Integral $int_0^fracpi2 x^2 sqrtsin x , dx$ @Frank Wei showed a way to evaluate your integral. I will try to show another method with the approach
I started the integral I posted there. $$I=int_0^fracpi2 x^2 sqrtcosxdx = int_0^fracpi2x^2 (1+tan^2 (x))^-frac14dx$$ substituting $tan x=y, $ we get $$ int_0^infty arctan^2 (x) (1+x^2)^-frac54 dx$$ $$I=-frac14int_0^inftylog^2left(frac1-ix1+ixright)(1+x^2)^-frac54 , dx$$ with $frac1-ix1+ix=y, $ we get the integral to be: $$I=Releft( - fraci 8sqrt 2int_-1^1 log^2 (y) , y^-frac54 sqrt1+y , dyright) $$Now we split the integral from $int_-1^0$ and $int_0^1 $substituting $y=-x$, the second one vanishes being purely imaginary, so we are left with:$$I=Re left(frac1+i16int_0^1 x^-frac54 sqrt1+x (-pi^2 +2 i pi log x +log^2 x) dxright) $$ $$I=-fracpi^216 int_0^1 x^-frac14-1(1-x)^frac32-1dx - fracpi 8int_0^1 x^-frac14-1(1-x)^frac32-1 log x,dx+frac1 16 int_0^1 x^-frac14-1(1-x)^frac32-1log^2(x),dx$$ We can evaluate these integrals using beta function, indeed $$I=-fracpi^2 16 Bleft(-frac14, frac32right) -fracpi 8fracddz Bleft(z, frac32right)big|_z=-frac14+frac1 16fracd^2 dz^2 Bleft(z, frac32right)big|_z=-frac14$$ I let you do the algebra in order to get $$I=sqrt 2pi^frac32fracpi^2+8G-16Gamma^2left(frac14right)$$
edited Jul 20 at 0:09
MathFacts
2,9921425
answered Jul 16 at 0:24
Zacky
2,2701327
1
Where $G$ is Catalan's constant, as per @FrankWei's answer
– peter a g
Jul 16 at 1:11
1
This is pure beauty ! Thanks for providing such an answer. Cheers.
– Claude Leibovici
Jul 16 at 5:19
Thank you for those kind words!
– Zacky
Jul 16 at 5:37
Cannot we use some other methods ? What if I split up $x^2$ as $-xleft(dfracpi2 -x right) + dfracpi x2$ ? I am wondering if that'll lead to a simple solution. Your solution is too advanced for me to understand.
– DarkKnight
Jul 16 at 6:12
It seems that $int_0^fracpi2 xsqrtcosx ,dx$ doesnt have a nice closed form so I doubt that would be easier. Surely there may be other approaches to this integral, have you seen @FrankWei's answer?
– Zacky
Jul 16 at 9:45
 |Â
show 1 more comment
up vote
12
down vote
up vote
12
down vote
In the question I asked here Integral $int_0^fracpi2 x^2 sqrtsin x , dx$ @Frank Wei showed a way to evaluate your integral. I will try to show another method with the approach
I started the integral I posted there. $$I=int_0^fracpi2 x^2 sqrtcosxdx = int_0^fracpi2x^2 (1+tan^2 (x))^-frac14dx$$ substituting $tan x=y, $ we get $$ int_0^infty arctan^2 (x) (1+x^2)^-frac54 dx$$ $$I=-frac14int_0^inftylog^2left(frac1-ix1+ixright)(1+x^2)^-frac54 , dx$$ with $frac1-ix1+ix=y, $ we get the integral to be: $$I=Releft( - fraci 8sqrt 2int_-1^1 log^2 (y) , y^-frac54 sqrt1+y , dyright) $$Now we split the integral from $int_-1^0$ and $int_0^1 $substituting $y=-x$, the second one vanishes being purely imaginary, so we are left with:$$I=Re left(frac1+i16int_0^1 x^-frac54 sqrt1+x (-pi^2 +2 i pi log x +log^2 x) dxright) $$ $$I=-fracpi^216 int_0^1 x^-frac14-1(1-x)^frac32-1dx - fracpi 8int_0^1 x^-frac14-1(1-x)^frac32-1 log x,dx+frac1 16 int_0^1 x^-frac14-1(1-x)^frac32-1log^2(x),dx$$ We can evaluate these integrals using beta function, indeed $$I=-fracpi^2 16 Bleft(-frac14, frac32right) -fracpi 8fracddz Bleft(z, frac32right)big|_z=-frac14+frac1 16fracd^2 dz^2 Bleft(z, frac32right)big|_z=-frac14$$ I let you do the algebra in order to get $$I=sqrt 2pi^frac32fracpi^2+8G-16Gamma^2left(frac14right)$$
edited Jul 20 at 0:09
MathFacts
2,9921425
answered Jul 16 at 0:24
Zacky
2,2701327
In the question I asked here Integral $int_0^fracpi2 x^2 sqrtsin x , dx$ @Frank Wei showed a way to evaluate your integral. I will try to show another method with the approach
I started the integral I posted there. $$I=int_0^fracpi2 x^2 sqrtcosxdx = int_0^fracpi2x^2 (1+tan^2 (x))^-frac14dx$$ substituting $tan x=y, $ we get $$ int_0^infty arctan^2 (x) (1+x^2)^-frac54 dx$$ $$I=-frac14int_0^inftylog^2left(frac1-ix1+ixright)(1+x^2)^-frac54 , dx$$ with $frac1-ix1+ix=y, $ we get the integral to be: $$I=Releft( - fraci 8sqrt 2int_-1^1 log^2 (y) , y^-frac54 sqrt1+y , dyright) $$Now we split the integral from $int_-1^0$ and $int_0^1 $substituting $y=-x$, the second one vanishes being purely imaginary, so we are left with:$$I=Re left(frac1+i16int_0^1 x^-frac54 sqrt1+x (-pi^2 +2 i pi log x +log^2 x) dxright) $$ $$I=-fracpi^216 int_0^1 x^-frac14-1(1-x)^frac32-1dx - fracpi 8int_0^1 x^-frac14-1(1-x)^frac32-1 log x,dx+frac1 16 int_0^1 x^-frac14-1(1-x)^frac32-1log^2(x),dx$$ We can evaluate these integrals using beta function, indeed $$I=-fracpi^2 16 Bleft(-frac14, frac32right) -fracpi 8fracddz Bleft(z, frac32right)big|_z=-frac14+frac1 16fracd^2 dz^2 Bleft(z, frac32right)big|_z=-frac14$$ I let you do the algebra in order to get $$I=sqrt 2pi^frac32fracpi^2+8G-16Gamma^2left(frac14right)$$
edited Jul 20 at 0:09
MathFacts
2,9921425
answered Jul 16 at 0:24
Zacky
2,2701327
edited Jul 20 at 0:09
MathFacts
2,9921425
edited Jul 20 at 0:09
MathFacts
2,9921425
edited Jul 20 at 0:09
MathFacts
2,9921425
2,9921425
answered Jul 16 at 0:24
Zacky
2,2701327
answered Jul 16 at 0:24
Zacky
2,2701327
answered Jul 16 at 0:24
Zacky
2,2701327
2,2701327
1
Where $G$ is Catalan's constant, as per @FrankWei's answer
– peter a g
Jul 16 at 1:11
1
This is pure beauty ! Thanks for providing such an answer. Cheers.
– Claude Leibovici
Jul 16 at 5:19
Thank you for those kind words!
– Zacky
Jul 16 at 5:37
Cannot we use some other methods ? What if I split up $x^2$ as $-xleft(dfracpi2 -x right) + dfracpi x2$ ? I am wondering if that'll lead to a simple solution. Your solution is too advanced for me to understand.
– DarkKnight
Jul 16 at 6:12
It seems that $int_0^fracpi2 xsqrtcosx ,dx$ doesnt have a nice closed form so I doubt that would be easier. Surely there may be other approaches to this integral, have you seen @FrankWei's answer?
– Zacky
Jul 16 at 9:45
 |Â
show 1 more comment
1
Where $G$ is Catalan's constant, as per @FrankWei's answer
– peter a g
Jul 16 at 1:11
1
This is pure beauty ! Thanks for providing such an answer. Cheers.
– Claude Leibovici
Jul 16 at 5:19
Thank you for those kind words!
– Zacky
Jul 16 at 5:37
Cannot we use some other methods ? What if I split up $x^2$ as $-xleft(dfracpi2 -x right) + dfracpi x2$ ? I am wondering if that'll lead to a simple solution. Your solution is too advanced for me to understand.
– DarkKnight
Jul 16 at 6:12
It seems that $int_0^fracpi2 xsqrtcosx ,dx$ doesnt have a nice closed form so I doubt that would be easier. Surely there may be other approaches to this integral, have you seen @FrankWei's answer?
– Zacky
Jul 16 at 9:45
1
1
Where $G$ is Catalan's constant, as per @FrankWei's answer
– peter a g
Jul 16 at 1:11
Where $G$ is Catalan's constant, as per @FrankWei's answer
– peter a g
Jul 16 at 1:11
1
1
This is pure beauty ! Thanks for providing such an answer. Cheers.
– Claude Leibovici
Jul 16 at 5:19
This is pure beauty ! Thanks for providing such an answer. Cheers.
– Claude Leibovici
Jul 16 at 5:19
Thank you for those kind words!
– Zacky
Jul 16 at 5:37
Thank you for those kind words!
– Zacky
Jul 16 at 5:37
Cannot we use some other methods ? What if I split up $x^2$ as $-xleft(dfracpi2 -x right) + dfracpi x2$ ? I am wondering if that'll lead to a simple solution. Your solution is too advanced for me to understand.
– DarkKnight
Jul 16 at 6:12
Cannot we use some other methods ? What if I split up $x^2$ as $-xleft(dfracpi2 -x right) + dfracpi x2$ ? I am wondering if that'll lead to a simple solution. Your solution is too advanced for me to understand.
– DarkKnight
Jul 16 at 6:12
It seems that $int_0^fracpi2 xsqrtcosx ,dx$ doesnt have a nice closed form so I doubt that would be easier. Surely there may be other approaches to this integral, have you seen @FrankWei's answer?
– Zacky
Jul 16 at 9:45
It seems that $int_0^fracpi2 xsqrtcosx ,dx$ doesnt have a nice closed form so I doubt that would be easier. Surely there may be other approaches to this integral, have you seen @FrankWei's answer?
– Zacky
Jul 16 at 9:45
 |Â
show 1 more comment
up vote
5
down vote
Just for the fun of it !
Since Dahaka provided a splendid and exact solution, I have been wondering what would give the approximation
$$cos(x) simeqfracpi ^2-4x^2pi ^2+x^2qquad (-frac pi 2 leq xleqfrac pi 2)$$ So, considering
$$I=int_0^frac pi 2x^2 sqrtfracpi ^2-4 x^2x^2+pi ^2,dx$$ the antiderivative expresses in terms of elliptic integrals but, using bounds, we get the simple
$$I=frac pi ^36 left(9 Eleft(-frac14right)-10
Kleft(-frac14right)right)approx 0.71832$$ which is in error by $0.14$%.
answered Jul 16 at 5:18
Claude Leibovici
112k1055126
add a comment |Â
up vote
5
down vote
Just for the fun of it !
Since Dahaka provided a splendid and exact solution, I have been wondering what would give the approximation
$$cos(x) simeqfracpi ^2-4x^2pi ^2+x^2qquad (-frac pi 2 leq xleqfrac pi 2)$$ So, considering
$$I=int_0^frac pi 2x^2 sqrtfracpi ^2-4 x^2x^2+pi ^2,dx$$ the antiderivative expresses in terms of elliptic integrals but, using bounds, we get the simple
$$I=frac pi ^36 left(9 Eleft(-frac14right)-10
Kleft(-frac14right)right)approx 0.71832$$ which is in error by $0.14$%.
answered Jul 16 at 5:18
Claude Leibovici
112k1055126
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Just for the fun of it !
Since Dahaka provided a splendid and exact solution, I have been wondering what would give the approximation
$$cos(x) simeqfracpi ^2-4x^2pi ^2+x^2qquad (-frac pi 2 leq xleqfrac pi 2)$$ So, considering
$$I=int_0^frac pi 2x^2 sqrtfracpi ^2-4 x^2x^2+pi ^2,dx$$ the antiderivative expresses in terms of elliptic integrals but, using bounds, we get the simple
$$I=frac pi ^36 left(9 Eleft(-frac14right)-10
Kleft(-frac14right)right)approx 0.71832$$ which is in error by $0.14$%.
answered Jul 16 at 5:18
Claude Leibovici
112k1055126
Just for the fun of it !
Since Dahaka provided a splendid and exact solution, I have been wondering what would give the approximation
$$cos(x) simeqfracpi ^2-4x^2pi ^2+x^2qquad (-frac pi 2 leq xleqfrac pi 2)$$ So, considering
$$I=int_0^frac pi 2x^2 sqrtfracpi ^2-4 x^2x^2+pi ^2,dx$$ the antiderivative expresses in terms of elliptic integrals but, using bounds, we get the simple
$$I=frac pi ^36 left(9 Eleft(-frac14right)-10
Kleft(-frac14right)right)approx 0.71832$$ which is in error by $0.14$%.
answered Jul 16 at 5:18
Claude Leibovici
112k1055126
answered Jul 16 at 5:18
Claude Leibovici
112k1055126
answered Jul 16 at 5:18
Claude Leibovici
112k1055126
answered Jul 16 at 5:18
Claude Leibovici
112k1055126
112k1055126
add a comment |Â
add a comment |Â
up vote
2
down vote
To find a primitive is arduous with the usual methods.
We could provide a rough approximation with the midpoint formula. Let $f(x)=x^2sqrtcos x$ $$int_0^pi/2f(x)approx fracpi2fbigg(fracpi4bigg)=0.8148.$$
If you want a more accurate approximation, a simple way is through the Legendre polynomials. The Legendre polynomias are given by $$P_n(x)=frac12^nn!fracd^ndx^n(x^2-1)^n.$$ We place $n=2$, $P_2(x)=frac12(3x^2-1)$. The quadrature formula, in this case, is
$$
int_-1^1f(x);dxapproxsum_i=1^2 A_if(x_i),
$$
where $$A_i=int_-1^1fracP_2(x)(x-x_i)P'_2(x_i);dx$$
and $x_i$ for $i=1,2$ are $P_2(x)$ roots.
Now, $A_1=A_2=1$ and $x_1=-frac1sqrt3$, $x_2=frac1sqrt3$.
Therefore
$$
int_a^b f(x);dx=fracb-a2int_-1^1 fbigg(fraca+b2+tfracb-a2bigg);dtapprox fracb-a2bigg[1cdot f(y_1)+1cdot f(y_2)bigg],
$$
where
$$
y_1=fraca+b2-frac1sqrt3fracb-a2quad textandquad y_2=fraca+b2+frac1sqrt3fracb-a2.
$$
Then
$$int_0^pi/2 f(x);dxapprox 0.77$$
answered Jul 15 at 22:58
Jack J.
3661317
add a comment |Â
up vote
2
down vote
To find a primitive is arduous with the usual methods.
We could provide a rough approximation with the midpoint formula. Let $f(x)=x^2sqrtcos x$ $$int_0^pi/2f(x)approx fracpi2fbigg(fracpi4bigg)=0.8148.$$
If you want a more accurate approximation, a simple way is through the Legendre polynomials. The Legendre polynomias are given by $$P_n(x)=frac12^nn!fracd^ndx^n(x^2-1)^n.$$ We place $n=2$, $P_2(x)=frac12(3x^2-1)$. The quadrature formula, in this case, is
$$
int_-1^1f(x);dxapproxsum_i=1^2 A_if(x_i),
$$
where $$A_i=int_-1^1fracP_2(x)(x-x_i)P'_2(x_i);dx$$
and $x_i$ for $i=1,2$ are $P_2(x)$ roots.
Now, $A_1=A_2=1$ and $x_1=-frac1sqrt3$, $x_2=frac1sqrt3$.
Therefore
$$
int_a^b f(x);dx=fracb-a2int_-1^1 fbigg(fraca+b2+tfracb-a2bigg);dtapprox fracb-a2bigg[1cdot f(y_1)+1cdot f(y_2)bigg],
$$
where
$$
y_1=fraca+b2-frac1sqrt3fracb-a2quad textandquad y_2=fraca+b2+frac1sqrt3fracb-a2.
$$
Then
$$int_0^pi/2 f(x);dxapprox 0.77$$
answered Jul 15 at 22:58
Jack J.
3661317
add a comment |Â
up vote
2
down vote
up vote
2
down vote
To find a primitive is arduous with the usual methods.
We could provide a rough approximation with the midpoint formula. Let $f(x)=x^2sqrtcos x$ $$int_0^pi/2f(x)approx fracpi2fbigg(fracpi4bigg)=0.8148.$$
If you want a more accurate approximation, a simple way is through the Legendre polynomials. The Legendre polynomias are given by $$P_n(x)=frac12^nn!fracd^ndx^n(x^2-1)^n.$$ We place $n=2$, $P_2(x)=frac12(3x^2-1)$. The quadrature formula, in this case, is
$$
int_-1^1f(x);dxapproxsum_i=1^2 A_if(x_i),
$$
where $$A_i=int_-1^1fracP_2(x)(x-x_i)P'_2(x_i);dx$$
and $x_i$ for $i=1,2$ are $P_2(x)$ roots.
Now, $A_1=A_2=1$ and $x_1=-frac1sqrt3$, $x_2=frac1sqrt3$.
Therefore
$$
int_a^b f(x);dx=fracb-a2int_-1^1 fbigg(fraca+b2+tfracb-a2bigg);dtapprox fracb-a2bigg[1cdot f(y_1)+1cdot f(y_2)bigg],
$$
where
$$
y_1=fraca+b2-frac1sqrt3fracb-a2quad textandquad y_2=fraca+b2+frac1sqrt3fracb-a2.
$$
Then
$$int_0^pi/2 f(x);dxapprox 0.77$$
answered Jul 15 at 22:58
Jack J.
3661317
To find a primitive is arduous with the usual methods.
We could provide a rough approximation with the midpoint formula. Let $f(x)=x^2sqrtcos x$ $$int_0^pi/2f(x)approx fracpi2fbigg(fracpi4bigg)=0.8148.$$
If you want a more accurate approximation, a simple way is through the Legendre polynomials. The Legendre polynomias are given by $$P_n(x)=frac12^nn!fracd^ndx^n(x^2-1)^n.$$ We place $n=2$, $P_2(x)=frac12(3x^2-1)$. The quadrature formula, in this case, is
$$
int_-1^1f(x);dxapproxsum_i=1^2 A_if(x_i),
$$
where $$A_i=int_-1^1fracP_2(x)(x-x_i)P'_2(x_i);dx$$
and $x_i$ for $i=1,2$ are $P_2(x)$ roots.
Now, $A_1=A_2=1$ and $x_1=-frac1sqrt3$, $x_2=frac1sqrt3$.
Therefore
$$
int_a^b f(x);dx=fracb-a2int_-1^1 fbigg(fraca+b2+tfracb-a2bigg);dtapprox fracb-a2bigg[1cdot f(y_1)+1cdot f(y_2)bigg],
$$
where
$$
y_1=fraca+b2-frac1sqrt3fracb-a2quad textandquad y_2=fraca+b2+frac1sqrt3fracb-a2.
$$
Then
$$int_0^pi/2 f(x);dxapprox 0.77$$
answered Jul 15 at 22:58
Jack J.
3661317
edited Jul 16 at 0:26
answered Jul 15 at 22:58
Jack J.
3661317
answered Jul 15 at 22:58
Jack J.
3661317
answered Jul 15 at 22:58
Jack J.
3661317
3661317
add a comment |Â
add a comment |Â
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1
I don't think that you will be able to find the antiderivative of $x^2sqrtcos(x)$: wolframAlpha gives that "monster"
– Holo
Jul 15 at 22:50
Where is the monster? An result containing the known elementary functions is impossible.
– Dr. Sonnhard Graubner
Jul 15 at 22:52
1
You can use a numerical way to catch this.
– Dr. Sonnhard Graubner
Jul 15 at 22:53
I want a certain value of integral.
– MathFacts
Jul 15 at 22:55
Here it is $$.71728411711845890580$$
– Dr. Sonnhard Graubner
Jul 15 at 23:01