Integrate $intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x$Existence and Uniqueness of Poisson Equation with Robin Boundary Condition using First Variation MethodsHow to integrate $int frac1cos(x),mathrm dx$Evaluating $int fracsinx+cosxsin^4x+cos^4x mathrm dx$Evaluating: $int 3xsinleft(frac x4right) , dx$.Shorter way to integrate $int fracx^9(x^2+4)^6 , mathrmdx$Integrate $intfracdxxsqrtx^2+1$How to integrate: $int fracsec xsqrtsin(2x + A) + sin A dx$?Integrate with substitution. Evaluate $ int fracsinsqrt xsqrt x dx$.Integrate $intfrac1xsqrtfrac1-x^21+x^2,mathrmdx$Integrate $intfrac1xsqrt1-x^3dx$Evaluating $int xsin^-1x dx$
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Integrate $intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x$
Existence and Uniqueness of Poisson Equation with Robin Boundary Condition using First Variation MethodsHow to integrate $int frac1cos(x),mathrm dx$Evaluating $int fracsinx+cosxsin^4x+cos^4x mathrm dx$Evaluating: $int 3xsinleft(frac x4right) , dx$.Shorter way to integrate $int fracx^9(x^2+4)^6 , mathrmdx$Integrate $intfracdxxsqrtx^2+1$How to integrate: $int fracsec xsqrtsin(2x + A) + sin A dx$?Integrate with substitution. Evaluate $ int fracsinsqrt xsqrt x dx$.Integrate $intfrac1xsqrtfrac1-x^21+x^2,mathrmdx$Integrate $intfrac1xsqrt1-x^3dx$Evaluating $int xsin^-1x dx$
$begingroup$
Integrate $$intfracsin^-1 (x)(1-x^2)^frac34 ,mathrm d x$$
I have followed some steps from here, but am not able to solve this question. Any help would be appreciated.
Update: After applying different logics, I somehow landed here:
$$intfrac1sqrt1-x^4$$
calculus integration trigonometry indefinite-integrals
edited Mar 17 at 22:58
Peiffap
10312
asked Mar 17 at 12:27
SanyaSanya
236
$endgroup$
|
show 7 more comments
$begingroup$
Integrate $$intfracsin^-1 (x)(1-x^2)^frac34 ,mathrm d x$$
I have followed some steps from here, but am not able to solve this question. Any help would be appreciated.
Update: After applying different logics, I somehow landed here:
$$intfrac1sqrt1-x^4$$
calculus integration trigonometry indefinite-integrals
edited Mar 17 at 22:58
Peiffap
10312
asked Mar 17 at 12:27
SanyaSanya
236
$endgroup$
$begingroup$
Welcome to MSE. Please show us the steps that you have taken.
$endgroup$
– José Carlos Santos
Mar 17 at 12:34
1
$begingroup$
Did you mean to change the exponent from $3/2$ in the linked source to $3/4$? Because the consequences are horrible.
$endgroup$
– J.G.
Mar 17 at 12:59
$begingroup$
Yes. This is the question (with 3/4) and the link is for reference. I have used some steps (basic approach) from that solution but couldn't solve this.
$endgroup$
– Sanya
Mar 17 at 13:02
$begingroup$
When you write $sin^-1(x)$, do you mean $arcsin(x)$ or $frac1sin(x)$?
$endgroup$
– Jon due
Mar 17 at 14:32
$begingroup$
@Jondue arcsin(x) or the inverse function of sinus
$endgroup$
– Sanya
Mar 17 at 14:40
|
show 7 more comments
$begingroup$
Integrate $$intfracsin^-1 (x)(1-x^2)^frac34 ,mathrm d x$$
I have followed some steps from here, but am not able to solve this question. Any help would be appreciated.
Update: After applying different logics, I somehow landed here:
$$intfrac1sqrt1-x^4$$
calculus integration trigonometry indefinite-integrals
edited Mar 17 at 22:58
Peiffap
10312
asked Mar 17 at 12:27
SanyaSanya
236
$endgroup$
Integrate $$intfracsin^-1 (x)(1-x^2)^frac34 ,mathrm d x$$
I have followed some steps from here, but am not able to solve this question. Any help would be appreciated.
Update: After applying different logics, I somehow landed here:
$$intfrac1sqrt1-x^4$$
calculus integration trigonometry indefinite-integrals
calculus integration trigonometry indefinite-integrals
edited Mar 17 at 22:58
Peiffap
10312
asked Mar 17 at 12:27
SanyaSanya
236
edited Mar 17 at 22:58
Peiffap
10312
asked Mar 17 at 12:27
SanyaSanya
236
edited Mar 17 at 22:58
Peiffap
10312
edited Mar 17 at 22:58
Peiffap
10312
edited Mar 17 at 22:58
Peiffap
10312
10312
asked Mar 17 at 12:27
SanyaSanya
236
asked Mar 17 at 12:27
SanyaSanya
236
asked Mar 17 at 12:27
SanyaSanya
236
236
$begingroup$
Welcome to MSE. Please show us the steps that you have taken.
$endgroup$
– José Carlos Santos
Mar 17 at 12:34
1
$begingroup$
Did you mean to change the exponent from $3/2$ in the linked source to $3/4$? Because the consequences are horrible.
$endgroup$
– J.G.
Mar 17 at 12:59
$begingroup$
Yes. This is the question (with 3/4) and the link is for reference. I have used some steps (basic approach) from that solution but couldn't solve this.
$endgroup$
– Sanya
Mar 17 at 13:02
$begingroup$
When you write $sin^-1(x)$, do you mean $arcsin(x)$ or $frac1sin(x)$?
$endgroup$
– Jon due
Mar 17 at 14:32
$begingroup$
@Jondue arcsin(x) or the inverse function of sinus
$endgroup$
– Sanya
Mar 17 at 14:40
|
show 7 more comments
$begingroup$
Welcome to MSE. Please show us the steps that you have taken.
$endgroup$
– José Carlos Santos
Mar 17 at 12:34
1
$begingroup$
Did you mean to change the exponent from $3/2$ in the linked source to $3/4$? Because the consequences are horrible.
$endgroup$
– J.G.
Mar 17 at 12:59
$begingroup$
Yes. This is the question (with 3/4) and the link is for reference. I have used some steps (basic approach) from that solution but couldn't solve this.
$endgroup$
– Sanya
Mar 17 at 13:02
$begingroup$
When you write $sin^-1(x)$, do you mean $arcsin(x)$ or $frac1sin(x)$?
$endgroup$
– Jon due
Mar 17 at 14:32
$begingroup$
@Jondue arcsin(x) or the inverse function of sinus
$endgroup$
– Sanya
Mar 17 at 14:40
$begingroup$
Welcome to MSE. Please show us the steps that you have taken.
$endgroup$
– José Carlos Santos
Mar 17 at 12:34
$begingroup$
Welcome to MSE. Please show us the steps that you have taken.
$endgroup$
– José Carlos Santos
Mar 17 at 12:34
1
1
$begingroup$
Did you mean to change the exponent from $3/2$ in the linked source to $3/4$? Because the consequences are horrible.
$endgroup$
– J.G.
Mar 17 at 12:59
$begingroup$
Did you mean to change the exponent from $3/2$ in the linked source to $3/4$? Because the consequences are horrible.
$endgroup$
– J.G.
Mar 17 at 12:59
$begingroup$
Yes. This is the question (with 3/4) and the link is for reference. I have used some steps (basic approach) from that solution but couldn't solve this.
$endgroup$
– Sanya
Mar 17 at 13:02
$begingroup$
Yes. This is the question (with 3/4) and the link is for reference. I have used some steps (basic approach) from that solution but couldn't solve this.
$endgroup$
– Sanya
Mar 17 at 13:02
$begingroup$
When you write $sin^-1(x)$, do you mean $arcsin(x)$ or $frac1sin(x)$?
$endgroup$
– Jon due
Mar 17 at 14:32
$begingroup$
When you write $sin^-1(x)$, do you mean $arcsin(x)$ or $frac1sin(x)$?
$endgroup$
– Jon due
Mar 17 at 14:32
$begingroup$
@Jondue arcsin(x) or the inverse function of sinus
$endgroup$
– Sanya
Mar 17 at 14:40
$begingroup$
@Jondue arcsin(x) or the inverse function of sinus
$endgroup$
– Sanya
Mar 17 at 14:40
|
show 7 more comments
3 Answers
3
active
oldest
votes
$begingroup$
If the solution in terms of hypergeometric functions
$$I=intfracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$ $$I=-frac12 sqrt[4]1-x^2 left(fracpi sqrt1-x^2 ,
_3F_2left(frac34,frac34,1;frac54,frac74;1-x^2right)sqrt
2 Gamma left(frac54right) Gamma left(frac74right)+4 x ,
_2F_1left(frac34,1;frac54;1-x^2right) sin ^-1(x)right)$$ is considered to be not acceptable, then approximations are required.
To my humble opinion, the most promising comes from @Carlos E. González C.'s answer
$$x=sin(y) implies I=intfracysqrtcos (y),mathrmdy$$
Expanding $frac1sqrtcos (y)$ as a Taylor series built at $y=0$,we end with
$$fracysqrtcos (y)=y+fracy^34+frac7 y^596+frac139 y^75760+frac5473
y^9645120+frac51103 y^1116588800+frac34988647
y^1330656102400+Oleft(y^15right)$$ Integrating termwise, then
$$I=fracy^22+fracy^416+frac7 y^6576+frac139 y^846080+frac5473
y^106451200+frac51103 y^12199065600+frac34988647
y^14429185433600+Oleft(y^16right)$$ where $y=sin^-1 (x)$.
Now, a few values for $$J=int_0 ^afracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$
$$left(
beginarrayccc
a & textapproximation & textexact \
0.05 & 0.0012514 & 0.0012514 \
0.10 & 0.0050231 & 0.0050231 \
0.15 & 0.0113677 & 0.0113677 \
0.20 & 0.0203761 & 0.0203761 \
0.25 & 0.0321816 & 0.0321817 \
0.30 & 0.0469674 & 0.0469674 \
0.35 & 0.0649765 & 0.0649765 \
0.40 & 0.0865271 & 0.0865271 \
0.45 & 0.1120350 & 0.1120346 \
0.50 & 0.1420440 & 0.1420443 \
0.55 & 0.1772810 & 0.1772811 \
0.60 & 0.2187270 & 0.2187274 \
0.65 & 0.2677510 & 0.2677515 \
0.70 & 0.3263300 & 0.3263310 \
0.75 & 0.3974690 & 0.3974711 \
0.80 & 0.4860560 & 0.4860678 \
0.85 & 0.6009030 & 0.6009565 \
0.90 & 0.7606450 & 0.7609314 \
0.95 & 1.0188000 & 1.0209564
endarray
right)$$
edited Mar 18 at 9:47
Peiffap
10312
answered Mar 18 at 5:49
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
It might be interesting to show how you arrived at the first expression using hypergeometric functions, even if it isn't considered acceptable.
$endgroup$
– Peiffap
Mar 18 at 9:08
1
$begingroup$
@Peiffap. This is the answer given by WA (see GuterBraten's answer).
$endgroup$
– Claude Leibovici
Mar 18 at 9:11
$begingroup$
I figured, but I feel like there's probably still value in finding how one can derive that answer and posting it on this site.
$endgroup$
– Peiffap
Mar 18 at 9:13
2
$begingroup$
@Peiffap. I agree with you but .... By the way, we could have something in the middle, still arriving to hypergeometric functions. Consider that $$frac1(1-x^2)^frac34=sum_p=0^infty fracGamma left(p+frac34right)Gamma left(frac34right) Gamma(p+1)x^2p $$ making that we are left with $$int sin^-1(x), x^2p,dx=fracx^2 n+1 left(2 (n+1) sin ^-1(x)-x , _2F_1left(frac12,n+1;n+2;x^2right)right)2 (n+1) (2 n+1)$$
$endgroup$
– Claude Leibovici
Mar 18 at 9:30
add a comment |
$begingroup$
I have tried to find an exact solution by Mathematica without any great success(https://www.wolframalpha.com/input/?i=arcsinx%2F(1-x%5E2)%5E(3%2F4)+integral).
Although there is an exact solution it dosen't seem resonable to evaluate it manually.
You could instead consider approximating the integral, which is possible by Taylor expansion.
I have made an relatively rough approximation here:
$intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x ≈ int(4x + frac383x^3) dx = frac19x^46+2x^2+C$
$intfracsin^-1 (x)(1-x^2)^3/4 ≈ frac19x^46+2x^2+C$
answered Mar 17 at 21:36
GuterBratenGuterBraten
428
$endgroup$
$begingroup$
I think that there is mistake in the Taylor expansion since, if I am not mistaken, the integrand is $x+frac11 x^312+Oleft(x^4right)$ making the integral to be $fracx^22+frac11 x^448+Oleft(x^5right)$. Using is for an upper bounf equal to $frac 12$, this would give $frac107768approx 0.139323$ which is not bad compared to the value in the table I produced.
$endgroup$
– Claude Leibovici
Mar 18 at 11:34
$begingroup$
I might have made an error or it is bescause i only made an approximation to order 3
$endgroup$
– GuterBraten
Mar 18 at 12:56
add a comment |
$begingroup$
maybe this can help you:
For clarity I'll change the notation, $sin^-1(x)=mathrmarcsin(x)$.
beginequation
int fracmathrmarcsin(x)(1-x^2)^3/4 dx
endequation
begineqnarray
y=mathrmarcsin(x) &quad& x=sin(y)\
dy=frac1(1-x^2)^1/2dx &quad& dx= (1-x^2)^1/2dy
endeqnarray
beginequation
int fracmathrmarcsin(x)(1-x^2)^3/4 dx = int fracmathrmarcsin(x)(1-x^2)^1/4(1-x^2)^1/2 dx
endequation
Applying the change of variables
beginequation
int fracy(1-sin^2(y))^1/4 dy = int fracysqrtcos(y)dy
endequation
edited Mar 17 at 22:15
eyeballfrog
6,904632
answered Mar 17 at 21:37
Carlos E. González C.Carlos E. González C.
605
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the solution in terms of hypergeometric functions
$$I=intfracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$ $$I=-frac12 sqrt[4]1-x^2 left(fracpi sqrt1-x^2 ,
_3F_2left(frac34,frac34,1;frac54,frac74;1-x^2right)sqrt
2 Gamma left(frac54right) Gamma left(frac74right)+4 x ,
_2F_1left(frac34,1;frac54;1-x^2right) sin ^-1(x)right)$$ is considered to be not acceptable, then approximations are required.
To my humble opinion, the most promising comes from @Carlos E. González C.'s answer
$$x=sin(y) implies I=intfracysqrtcos (y),mathrmdy$$
Expanding $frac1sqrtcos (y)$ as a Taylor series built at $y=0$,we end with
$$fracysqrtcos (y)=y+fracy^34+frac7 y^596+frac139 y^75760+frac5473
y^9645120+frac51103 y^1116588800+frac34988647
y^1330656102400+Oleft(y^15right)$$ Integrating termwise, then
$$I=fracy^22+fracy^416+frac7 y^6576+frac139 y^846080+frac5473
y^106451200+frac51103 y^12199065600+frac34988647
y^14429185433600+Oleft(y^16right)$$ where $y=sin^-1 (x)$.
Now, a few values for $$J=int_0 ^afracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$
$$left(
beginarrayccc
a & textapproximation & textexact \
0.05 & 0.0012514 & 0.0012514 \
0.10 & 0.0050231 & 0.0050231 \
0.15 & 0.0113677 & 0.0113677 \
0.20 & 0.0203761 & 0.0203761 \
0.25 & 0.0321816 & 0.0321817 \
0.30 & 0.0469674 & 0.0469674 \
0.35 & 0.0649765 & 0.0649765 \
0.40 & 0.0865271 & 0.0865271 \
0.45 & 0.1120350 & 0.1120346 \
0.50 & 0.1420440 & 0.1420443 \
0.55 & 0.1772810 & 0.1772811 \
0.60 & 0.2187270 & 0.2187274 \
0.65 & 0.2677510 & 0.2677515 \
0.70 & 0.3263300 & 0.3263310 \
0.75 & 0.3974690 & 0.3974711 \
0.80 & 0.4860560 & 0.4860678 \
0.85 & 0.6009030 & 0.6009565 \
0.90 & 0.7606450 & 0.7609314 \
0.95 & 1.0188000 & 1.0209564
endarray
right)$$
edited Mar 18 at 9:47
Peiffap
10312
answered Mar 18 at 5:49
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
It might be interesting to show how you arrived at the first expression using hypergeometric functions, even if it isn't considered acceptable.
$endgroup$
– Peiffap
Mar 18 at 9:08
1
$begingroup$
@Peiffap. This is the answer given by WA (see GuterBraten's answer).
$endgroup$
– Claude Leibovici
Mar 18 at 9:11
$begingroup$
I figured, but I feel like there's probably still value in finding how one can derive that answer and posting it on this site.
$endgroup$
– Peiffap
Mar 18 at 9:13
2
$begingroup$
@Peiffap. I agree with you but .... By the way, we could have something in the middle, still arriving to hypergeometric functions. Consider that $$frac1(1-x^2)^frac34=sum_p=0^infty fracGamma left(p+frac34right)Gamma left(frac34right) Gamma(p+1)x^2p $$ making that we are left with $$int sin^-1(x), x^2p,dx=fracx^2 n+1 left(2 (n+1) sin ^-1(x)-x , _2F_1left(frac12,n+1;n+2;x^2right)right)2 (n+1) (2 n+1)$$
$endgroup$
– Claude Leibovici
Mar 18 at 9:30
add a comment |
$begingroup$
If the solution in terms of hypergeometric functions
$$I=intfracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$ $$I=-frac12 sqrt[4]1-x^2 left(fracpi sqrt1-x^2 ,
_3F_2left(frac34,frac34,1;frac54,frac74;1-x^2right)sqrt
2 Gamma left(frac54right) Gamma left(frac74right)+4 x ,
_2F_1left(frac34,1;frac54;1-x^2right) sin ^-1(x)right)$$ is considered to be not acceptable, then approximations are required.
To my humble opinion, the most promising comes from @Carlos E. González C.'s answer
$$x=sin(y) implies I=intfracysqrtcos (y),mathrmdy$$
Expanding $frac1sqrtcos (y)$ as a Taylor series built at $y=0$,we end with
$$fracysqrtcos (y)=y+fracy^34+frac7 y^596+frac139 y^75760+frac5473
y^9645120+frac51103 y^1116588800+frac34988647
y^1330656102400+Oleft(y^15right)$$ Integrating termwise, then
$$I=fracy^22+fracy^416+frac7 y^6576+frac139 y^846080+frac5473
y^106451200+frac51103 y^12199065600+frac34988647
y^14429185433600+Oleft(y^16right)$$ where $y=sin^-1 (x)$.
Now, a few values for $$J=int_0 ^afracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$
$$left(
beginarrayccc
a & textapproximation & textexact \
0.05 & 0.0012514 & 0.0012514 \
0.10 & 0.0050231 & 0.0050231 \
0.15 & 0.0113677 & 0.0113677 \
0.20 & 0.0203761 & 0.0203761 \
0.25 & 0.0321816 & 0.0321817 \
0.30 & 0.0469674 & 0.0469674 \
0.35 & 0.0649765 & 0.0649765 \
0.40 & 0.0865271 & 0.0865271 \
0.45 & 0.1120350 & 0.1120346 \
0.50 & 0.1420440 & 0.1420443 \
0.55 & 0.1772810 & 0.1772811 \
0.60 & 0.2187270 & 0.2187274 \
0.65 & 0.2677510 & 0.2677515 \
0.70 & 0.3263300 & 0.3263310 \
0.75 & 0.3974690 & 0.3974711 \
0.80 & 0.4860560 & 0.4860678 \
0.85 & 0.6009030 & 0.6009565 \
0.90 & 0.7606450 & 0.7609314 \
0.95 & 1.0188000 & 1.0209564
endarray
right)$$
edited Mar 18 at 9:47
Peiffap
10312
answered Mar 18 at 5:49
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
It might be interesting to show how you arrived at the first expression using hypergeometric functions, even if it isn't considered acceptable.
$endgroup$
– Peiffap
Mar 18 at 9:08
1
$begingroup$
@Peiffap. This is the answer given by WA (see GuterBraten's answer).
$endgroup$
– Claude Leibovici
Mar 18 at 9:11
$begingroup$
I figured, but I feel like there's probably still value in finding how one can derive that answer and posting it on this site.
$endgroup$
– Peiffap
Mar 18 at 9:13
2
$begingroup$
@Peiffap. I agree with you but .... By the way, we could have something in the middle, still arriving to hypergeometric functions. Consider that $$frac1(1-x^2)^frac34=sum_p=0^infty fracGamma left(p+frac34right)Gamma left(frac34right) Gamma(p+1)x^2p $$ making that we are left with $$int sin^-1(x), x^2p,dx=fracx^2 n+1 left(2 (n+1) sin ^-1(x)-x , _2F_1left(frac12,n+1;n+2;x^2right)right)2 (n+1) (2 n+1)$$
$endgroup$
– Claude Leibovici
Mar 18 at 9:30
add a comment |
$begingroup$
If the solution in terms of hypergeometric functions
$$I=intfracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$ $$I=-frac12 sqrt[4]1-x^2 left(fracpi sqrt1-x^2 ,
_3F_2left(frac34,frac34,1;frac54,frac74;1-x^2right)sqrt
2 Gamma left(frac54right) Gamma left(frac74right)+4 x ,
_2F_1left(frac34,1;frac54;1-x^2right) sin ^-1(x)right)$$ is considered to be not acceptable, then approximations are required.
To my humble opinion, the most promising comes from @Carlos E. González C.'s answer
$$x=sin(y) implies I=intfracysqrtcos (y),mathrmdy$$
Expanding $frac1sqrtcos (y)$ as a Taylor series built at $y=0$,we end with
$$fracysqrtcos (y)=y+fracy^34+frac7 y^596+frac139 y^75760+frac5473
y^9645120+frac51103 y^1116588800+frac34988647
y^1330656102400+Oleft(y^15right)$$ Integrating termwise, then
$$I=fracy^22+fracy^416+frac7 y^6576+frac139 y^846080+frac5473
y^106451200+frac51103 y^12199065600+frac34988647
y^14429185433600+Oleft(y^16right)$$ where $y=sin^-1 (x)$.
Now, a few values for $$J=int_0 ^afracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$
$$left(
beginarrayccc
a & textapproximation & textexact \
0.05 & 0.0012514 & 0.0012514 \
0.10 & 0.0050231 & 0.0050231 \
0.15 & 0.0113677 & 0.0113677 \
0.20 & 0.0203761 & 0.0203761 \
0.25 & 0.0321816 & 0.0321817 \
0.30 & 0.0469674 & 0.0469674 \
0.35 & 0.0649765 & 0.0649765 \
0.40 & 0.0865271 & 0.0865271 \
0.45 & 0.1120350 & 0.1120346 \
0.50 & 0.1420440 & 0.1420443 \
0.55 & 0.1772810 & 0.1772811 \
0.60 & 0.2187270 & 0.2187274 \
0.65 & 0.2677510 & 0.2677515 \
0.70 & 0.3263300 & 0.3263310 \
0.75 & 0.3974690 & 0.3974711 \
0.80 & 0.4860560 & 0.4860678 \
0.85 & 0.6009030 & 0.6009565 \
0.90 & 0.7606450 & 0.7609314 \
0.95 & 1.0188000 & 1.0209564
endarray
right)$$
edited Mar 18 at 9:47
Peiffap
10312
answered Mar 18 at 5:49
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
If the solution in terms of hypergeometric functions
$$I=intfracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$ $$I=-frac12 sqrt[4]1-x^2 left(fracpi sqrt1-x^2 ,
_3F_2left(frac34,frac34,1;frac54,frac74;1-x^2right)sqrt
2 Gamma left(frac54right) Gamma left(frac74right)+4 x ,
_2F_1left(frac34,1;frac54;1-x^2right) sin ^-1(x)right)$$ is considered to be not acceptable, then approximations are required.
To my humble opinion, the most promising comes from @Carlos E. González C.'s answer
$$x=sin(y) implies I=intfracysqrtcos (y),mathrmdy$$
Expanding $frac1sqrtcos (y)$ as a Taylor series built at $y=0$,we end with
$$fracysqrtcos (y)=y+fracy^34+frac7 y^596+frac139 y^75760+frac5473
y^9645120+frac51103 y^1116588800+frac34988647
y^1330656102400+Oleft(y^15right)$$ Integrating termwise, then
$$I=fracy^22+fracy^416+frac7 y^6576+frac139 y^846080+frac5473
y^106451200+frac51103 y^12199065600+frac34988647
y^14429185433600+Oleft(y^16right)$$ where $y=sin^-1 (x)$.
Now, a few values for $$J=int_0 ^afracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$
$$left(
beginarrayccc
a & textapproximation & textexact \
0.05 & 0.0012514 & 0.0012514 \
0.10 & 0.0050231 & 0.0050231 \
0.15 & 0.0113677 & 0.0113677 \
0.20 & 0.0203761 & 0.0203761 \
0.25 & 0.0321816 & 0.0321817 \
0.30 & 0.0469674 & 0.0469674 \
0.35 & 0.0649765 & 0.0649765 \
0.40 & 0.0865271 & 0.0865271 \
0.45 & 0.1120350 & 0.1120346 \
0.50 & 0.1420440 & 0.1420443 \
0.55 & 0.1772810 & 0.1772811 \
0.60 & 0.2187270 & 0.2187274 \
0.65 & 0.2677510 & 0.2677515 \
0.70 & 0.3263300 & 0.3263310 \
0.75 & 0.3974690 & 0.3974711 \
0.80 & 0.4860560 & 0.4860678 \
0.85 & 0.6009030 & 0.6009565 \
0.90 & 0.7606450 & 0.7609314 \
0.95 & 1.0188000 & 1.0209564
endarray
right)$$
edited Mar 18 at 9:47
Peiffap
10312
answered Mar 18 at 5:49
Claude LeiboviciClaude Leibovici
125k1158135
edited Mar 18 at 9:47
Peiffap
10312
edited Mar 18 at 9:47
Peiffap
10312
edited Mar 18 at 9:47
Peiffap
10312
10312
answered Mar 18 at 5:49
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 18 at 5:49
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 18 at 5:49
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
$begingroup$
It might be interesting to show how you arrived at the first expression using hypergeometric functions, even if it isn't considered acceptable.
$endgroup$
– Peiffap
Mar 18 at 9:08
1
$begingroup$
@Peiffap. This is the answer given by WA (see GuterBraten's answer).
$endgroup$
– Claude Leibovici
Mar 18 at 9:11
$begingroup$
I figured, but I feel like there's probably still value in finding how one can derive that answer and posting it on this site.
$endgroup$
– Peiffap
Mar 18 at 9:13
2
$begingroup$
@Peiffap. I agree with you but .... By the way, we could have something in the middle, still arriving to hypergeometric functions. Consider that $$frac1(1-x^2)^frac34=sum_p=0^infty fracGamma left(p+frac34right)Gamma left(frac34right) Gamma(p+1)x^2p $$ making that we are left with $$int sin^-1(x), x^2p,dx=fracx^2 n+1 left(2 (n+1) sin ^-1(x)-x , _2F_1left(frac12,n+1;n+2;x^2right)right)2 (n+1) (2 n+1)$$
$endgroup$
– Claude Leibovici
Mar 18 at 9:30
add a comment |
$begingroup$
It might be interesting to show how you arrived at the first expression using hypergeometric functions, even if it isn't considered acceptable.
$endgroup$
– Peiffap
Mar 18 at 9:08
1
$begingroup$
@Peiffap. This is the answer given by WA (see GuterBraten's answer).
$endgroup$
– Claude Leibovici
Mar 18 at 9:11
$begingroup$
I figured, but I feel like there's probably still value in finding how one can derive that answer and posting it on this site.
$endgroup$
– Peiffap
Mar 18 at 9:13
2
$begingroup$
@Peiffap. I agree with you but .... By the way, we could have something in the middle, still arriving to hypergeometric functions. Consider that $$frac1(1-x^2)^frac34=sum_p=0^infty fracGamma left(p+frac34right)Gamma left(frac34right) Gamma(p+1)x^2p $$ making that we are left with $$int sin^-1(x), x^2p,dx=fracx^2 n+1 left(2 (n+1) sin ^-1(x)-x , _2F_1left(frac12,n+1;n+2;x^2right)right)2 (n+1) (2 n+1)$$
$endgroup$
– Claude Leibovici
Mar 18 at 9:30
$begingroup$
It might be interesting to show how you arrived at the first expression using hypergeometric functions, even if it isn't considered acceptable.
$endgroup$
– Peiffap
Mar 18 at 9:08
$begingroup$
It might be interesting to show how you arrived at the first expression using hypergeometric functions, even if it isn't considered acceptable.
$endgroup$
– Peiffap
Mar 18 at 9:08
1
1
$begingroup$
@Peiffap. This is the answer given by WA (see GuterBraten's answer).
$endgroup$
– Claude Leibovici
Mar 18 at 9:11
$begingroup$
@Peiffap. This is the answer given by WA (see GuterBraten's answer).
$endgroup$
– Claude Leibovici
Mar 18 at 9:11
$begingroup$
I figured, but I feel like there's probably still value in finding how one can derive that answer and posting it on this site.
$endgroup$
– Peiffap
Mar 18 at 9:13
$begingroup$
I figured, but I feel like there's probably still value in finding how one can derive that answer and posting it on this site.
$endgroup$
– Peiffap
Mar 18 at 9:13
2
2
$begingroup$
@Peiffap. I agree with you but .... By the way, we could have something in the middle, still arriving to hypergeometric functions. Consider that $$frac1(1-x^2)^frac34=sum_p=0^infty fracGamma left(p+frac34right)Gamma left(frac34right) Gamma(p+1)x^2p $$ making that we are left with $$int sin^-1(x), x^2p,dx=fracx^2 n+1 left(2 (n+1) sin ^-1(x)-x , _2F_1left(frac12,n+1;n+2;x^2right)right)2 (n+1) (2 n+1)$$
$endgroup$
– Claude Leibovici
Mar 18 at 9:30
$begingroup$
@Peiffap. I agree with you but .... By the way, we could have something in the middle, still arriving to hypergeometric functions. Consider that $$frac1(1-x^2)^frac34=sum_p=0^infty fracGamma left(p+frac34right)Gamma left(frac34right) Gamma(p+1)x^2p $$ making that we are left with $$int sin^-1(x), x^2p,dx=fracx^2 n+1 left(2 (n+1) sin ^-1(x)-x , _2F_1left(frac12,n+1;n+2;x^2right)right)2 (n+1) (2 n+1)$$
$endgroup$
– Claude Leibovici
Mar 18 at 9:30
add a comment |
$begingroup$
I have tried to find an exact solution by Mathematica without any great success(https://www.wolframalpha.com/input/?i=arcsinx%2F(1-x%5E2)%5E(3%2F4)+integral).
Although there is an exact solution it dosen't seem resonable to evaluate it manually.
You could instead consider approximating the integral, which is possible by Taylor expansion.
I have made an relatively rough approximation here:
$intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x ≈ int(4x + frac383x^3) dx = frac19x^46+2x^2+C$
$intfracsin^-1 (x)(1-x^2)^3/4 ≈ frac19x^46+2x^2+C$
answered Mar 17 at 21:36
GuterBratenGuterBraten
428
$endgroup$
$begingroup$
I think that there is mistake in the Taylor expansion since, if I am not mistaken, the integrand is $x+frac11 x^312+Oleft(x^4right)$ making the integral to be $fracx^22+frac11 x^448+Oleft(x^5right)$. Using is for an upper bounf equal to $frac 12$, this would give $frac107768approx 0.139323$ which is not bad compared to the value in the table I produced.
$endgroup$
– Claude Leibovici
Mar 18 at 11:34
$begingroup$
I might have made an error or it is bescause i only made an approximation to order 3
$endgroup$
– GuterBraten
Mar 18 at 12:56
add a comment |
$begingroup$
I have tried to find an exact solution by Mathematica without any great success(https://www.wolframalpha.com/input/?i=arcsinx%2F(1-x%5E2)%5E(3%2F4)+integral).
Although there is an exact solution it dosen't seem resonable to evaluate it manually.
You could instead consider approximating the integral, which is possible by Taylor expansion.
I have made an relatively rough approximation here:
$intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x ≈ int(4x + frac383x^3) dx = frac19x^46+2x^2+C$
$intfracsin^-1 (x)(1-x^2)^3/4 ≈ frac19x^46+2x^2+C$
answered Mar 17 at 21:36
GuterBratenGuterBraten
428
$endgroup$
$begingroup$
I think that there is mistake in the Taylor expansion since, if I am not mistaken, the integrand is $x+frac11 x^312+Oleft(x^4right)$ making the integral to be $fracx^22+frac11 x^448+Oleft(x^5right)$. Using is for an upper bounf equal to $frac 12$, this would give $frac107768approx 0.139323$ which is not bad compared to the value in the table I produced.
$endgroup$
– Claude Leibovici
Mar 18 at 11:34
$begingroup$
I might have made an error or it is bescause i only made an approximation to order 3
$endgroup$
– GuterBraten
Mar 18 at 12:56
add a comment |
$begingroup$
I have tried to find an exact solution by Mathematica without any great success(https://www.wolframalpha.com/input/?i=arcsinx%2F(1-x%5E2)%5E(3%2F4)+integral).
Although there is an exact solution it dosen't seem resonable to evaluate it manually.
You could instead consider approximating the integral, which is possible by Taylor expansion.
I have made an relatively rough approximation here:
$intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x ≈ int(4x + frac383x^3) dx = frac19x^46+2x^2+C$
$intfracsin^-1 (x)(1-x^2)^3/4 ≈ frac19x^46+2x^2+C$
answered Mar 17 at 21:36
GuterBratenGuterBraten
428
$endgroup$
I have tried to find an exact solution by Mathematica without any great success(https://www.wolframalpha.com/input/?i=arcsinx%2F(1-x%5E2)%5E(3%2F4)+integral).
Although there is an exact solution it dosen't seem resonable to evaluate it manually.
You could instead consider approximating the integral, which is possible by Taylor expansion.
I have made an relatively rough approximation here:
$intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x ≈ int(4x + frac383x^3) dx = frac19x^46+2x^2+C$
$intfracsin^-1 (x)(1-x^2)^3/4 ≈ frac19x^46+2x^2+C$
answered Mar 17 at 21:36
GuterBratenGuterBraten
428
answered Mar 17 at 21:36
GuterBratenGuterBraten
428
answered Mar 17 at 21:36
GuterBratenGuterBraten
428
answered Mar 17 at 21:36
GuterBratenGuterBraten
428
428
$begingroup$
I think that there is mistake in the Taylor expansion since, if I am not mistaken, the integrand is $x+frac11 x^312+Oleft(x^4right)$ making the integral to be $fracx^22+frac11 x^448+Oleft(x^5right)$. Using is for an upper bounf equal to $frac 12$, this would give $frac107768approx 0.139323$ which is not bad compared to the value in the table I produced.
$endgroup$
– Claude Leibovici
Mar 18 at 11:34
$begingroup$
I might have made an error or it is bescause i only made an approximation to order 3
$endgroup$
– GuterBraten
Mar 18 at 12:56
add a comment |
$begingroup$
I think that there is mistake in the Taylor expansion since, if I am not mistaken, the integrand is $x+frac11 x^312+Oleft(x^4right)$ making the integral to be $fracx^22+frac11 x^448+Oleft(x^5right)$. Using is for an upper bounf equal to $frac 12$, this would give $frac107768approx 0.139323$ which is not bad compared to the value in the table I produced.
$endgroup$
– Claude Leibovici
Mar 18 at 11:34
$begingroup$
I might have made an error or it is bescause i only made an approximation to order 3
$endgroup$
– GuterBraten
Mar 18 at 12:56
$begingroup$
I think that there is mistake in the Taylor expansion since, if I am not mistaken, the integrand is $x+frac11 x^312+Oleft(x^4right)$ making the integral to be $fracx^22+frac11 x^448+Oleft(x^5right)$. Using is for an upper bounf equal to $frac 12$, this would give $frac107768approx 0.139323$ which is not bad compared to the value in the table I produced.
$endgroup$
– Claude Leibovici
Mar 18 at 11:34
$begingroup$
I think that there is mistake in the Taylor expansion since, if I am not mistaken, the integrand is $x+frac11 x^312+Oleft(x^4right)$ making the integral to be $fracx^22+frac11 x^448+Oleft(x^5right)$. Using is for an upper bounf equal to $frac 12$, this would give $frac107768approx 0.139323$ which is not bad compared to the value in the table I produced.
$endgroup$
– Claude Leibovici
Mar 18 at 11:34
$begingroup$
I might have made an error or it is bescause i only made an approximation to order 3
$endgroup$
– GuterBraten
Mar 18 at 12:56
$begingroup$
I might have made an error or it is bescause i only made an approximation to order 3
$endgroup$
– GuterBraten
Mar 18 at 12:56
add a comment |
$begingroup$
maybe this can help you:
For clarity I'll change the notation, $sin^-1(x)=mathrmarcsin(x)$.
beginequation
int fracmathrmarcsin(x)(1-x^2)^3/4 dx
endequation
begineqnarray
y=mathrmarcsin(x) &quad& x=sin(y)\
dy=frac1(1-x^2)^1/2dx &quad& dx= (1-x^2)^1/2dy
endeqnarray
beginequation
int fracmathrmarcsin(x)(1-x^2)^3/4 dx = int fracmathrmarcsin(x)(1-x^2)^1/4(1-x^2)^1/2 dx
endequation
Applying the change of variables
beginequation
int fracy(1-sin^2(y))^1/4 dy = int fracysqrtcos(y)dy
endequation
edited Mar 17 at 22:15
eyeballfrog
6,904632
answered Mar 17 at 21:37
Carlos E. González C.Carlos E. González C.
605
$endgroup$
add a comment |
$begingroup$
maybe this can help you:
For clarity I'll change the notation, $sin^-1(x)=mathrmarcsin(x)$.
beginequation
int fracmathrmarcsin(x)(1-x^2)^3/4 dx
endequation
begineqnarray
y=mathrmarcsin(x) &quad& x=sin(y)\
dy=frac1(1-x^2)^1/2dx &quad& dx= (1-x^2)^1/2dy
endeqnarray
beginequation
int fracmathrmarcsin(x)(1-x^2)^3/4 dx = int fracmathrmarcsin(x)(1-x^2)^1/4(1-x^2)^1/2 dx
endequation
Applying the change of variables
beginequation
int fracy(1-sin^2(y))^1/4 dy = int fracysqrtcos(y)dy
endequation
edited Mar 17 at 22:15
eyeballfrog
6,904632
answered Mar 17 at 21:37
Carlos E. González C.Carlos E. González C.
605
$endgroup$
add a comment |
$begingroup$
maybe this can help you:
For clarity I'll change the notation, $sin^-1(x)=mathrmarcsin(x)$.
beginequation
int fracmathrmarcsin(x)(1-x^2)^3/4 dx
endequation
begineqnarray
y=mathrmarcsin(x) &quad& x=sin(y)\
dy=frac1(1-x^2)^1/2dx &quad& dx= (1-x^2)^1/2dy
endeqnarray
beginequation
int fracmathrmarcsin(x)(1-x^2)^3/4 dx = int fracmathrmarcsin(x)(1-x^2)^1/4(1-x^2)^1/2 dx
endequation
Applying the change of variables
beginequation
int fracy(1-sin^2(y))^1/4 dy = int fracysqrtcos(y)dy
endequation
edited Mar 17 at 22:15
eyeballfrog
6,904632
answered Mar 17 at 21:37
Carlos E. González C.Carlos E. González C.
605
$endgroup$
maybe this can help you:
For clarity I'll change the notation, $sin^-1(x)=mathrmarcsin(x)$.
beginequation
int fracmathrmarcsin(x)(1-x^2)^3/4 dx
endequation
begineqnarray
y=mathrmarcsin(x) &quad& x=sin(y)\
dy=frac1(1-x^2)^1/2dx &quad& dx= (1-x^2)^1/2dy
endeqnarray
beginequation
int fracmathrmarcsin(x)(1-x^2)^3/4 dx = int fracmathrmarcsin(x)(1-x^2)^1/4(1-x^2)^1/2 dx
endequation
Applying the change of variables
beginequation
int fracy(1-sin^2(y))^1/4 dy = int fracysqrtcos(y)dy
endequation
edited Mar 17 at 22:15
eyeballfrog
6,904632
answered Mar 17 at 21:37
Carlos E. González C.Carlos E. González C.
605
edited Mar 17 at 22:15
eyeballfrog
6,904632
edited Mar 17 at 22:15
eyeballfrog
6,904632
edited Mar 17 at 22:15
eyeballfrog
6,904632
6,904632
answered Mar 17 at 21:37
Carlos E. González C.Carlos E. González C.
605
answered Mar 17 at 21:37
Carlos E. González C.Carlos E. González C.
605
answered Mar 17 at 21:37
Carlos E. González C.Carlos E. González C.
605
605
add a comment |
add a comment |
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Welcome to MSE. Please show us the steps that you have taken.
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– José Carlos Santos
Mar 17 at 12:34
1
$begingroup$
Did you mean to change the exponent from $3/2$ in the linked source to $3/4$? Because the consequences are horrible.
$endgroup$
– J.G.
Mar 17 at 12:59
$begingroup$
Yes. This is the question (with 3/4) and the link is for reference. I have used some steps (basic approach) from that solution but couldn't solve this.
$endgroup$
– Sanya
Mar 17 at 13:02
$begingroup$
When you write $sin^-1(x)$, do you mean $arcsin(x)$ or $frac1sin(x)$?
$endgroup$
– Jon due
Mar 17 at 14:32
$begingroup$
@Jondue arcsin(x) or the inverse function of sinus
$endgroup$
– Sanya
Mar 17 at 14:40