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 $int xsin^-1x dx$













2












$begingroup$



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







share|cite|improve this question











$endgroup$











  • $begingroup$
    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















2












$begingroup$



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







share|cite|improve this question











$endgroup$











  • $begingroup$
    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













2












2








2


1



$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 22:58









Peiffap

10312




10312










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
















  • $begingroup$
    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










3 Answers
3






active

oldest

votes


















4












$begingroup$

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 $$J=int_0 ^afracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$



$$left(
beginarrayccc
a & textapproximation & textexact \
0.05 & 0.0012514 & 0.0012514 \
0.10 & 0.0050231 & 0.0050231 \
0.15 & 0.0113677 & 0.0113677 \
0.20 & 0.0203761 & 0.0203761 \
0.25 & 0.0321816 & 0.0321817 \
0.30 & 0.0469674 & 0.0469674 \
0.35 & 0.0649765 & 0.0649765 \
0.40 & 0.0865271 & 0.0865271 \
0.45 & 0.1120350 & 0.1120346 \
0.50 & 0.1420440 & 0.1420443 \
0.55 & 0.1772810 & 0.1772811 \
0.60 & 0.2187270 & 0.2187274 \
0.65 & 0.2677510 & 0.2677515 \
0.70 & 0.3263300 & 0.3263310 \
0.75 & 0.3974690 & 0.3974711 \
0.80 & 0.4860560 & 0.4860678 \
0.85 & 0.6009030 & 0.6009565 \
0.90 & 0.7606450 & 0.7609314 \
0.95 & 1.0188000 & 1.0209564
endarray
right)$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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



















1












$begingroup$

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:



$intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x ≈ int(4x + frac383x^3) dx = frac19x^46+2x^2+C$



$intfracsin^-1 (x)(1-x^2)^3/4 ≈ frac19x^46+2x^2+C$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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


















1












$begingroup$

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






share|cite|improve this answer











$endgroup$












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    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 $$J=int_0 ^afracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$



    $$left(
    beginarrayccc
    a & textapproximation & textexact \
    0.05 & 0.0012514 & 0.0012514 \
    0.10 & 0.0050231 & 0.0050231 \
    0.15 & 0.0113677 & 0.0113677 \
    0.20 & 0.0203761 & 0.0203761 \
    0.25 & 0.0321816 & 0.0321817 \
    0.30 & 0.0469674 & 0.0469674 \
    0.35 & 0.0649765 & 0.0649765 \
    0.40 & 0.0865271 & 0.0865271 \
    0.45 & 0.1120350 & 0.1120346 \
    0.50 & 0.1420440 & 0.1420443 \
    0.55 & 0.1772810 & 0.1772811 \
    0.60 & 0.2187270 & 0.2187274 \
    0.65 & 0.2677510 & 0.2677515 \
    0.70 & 0.3263300 & 0.3263310 \
    0.75 & 0.3974690 & 0.3974711 \
    0.80 & 0.4860560 & 0.4860678 \
    0.85 & 0.6009030 & 0.6009565 \
    0.90 & 0.7606450 & 0.7609314 \
    0.95 & 1.0188000 & 1.0209564
    endarray
    right)$$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      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
















    4












    $begingroup$

    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 $$J=int_0 ^afracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$



    $$left(
    beginarrayccc
    a & textapproximation & textexact \
    0.05 & 0.0012514 & 0.0012514 \
    0.10 & 0.0050231 & 0.0050231 \
    0.15 & 0.0113677 & 0.0113677 \
    0.20 & 0.0203761 & 0.0203761 \
    0.25 & 0.0321816 & 0.0321817 \
    0.30 & 0.0469674 & 0.0469674 \
    0.35 & 0.0649765 & 0.0649765 \
    0.40 & 0.0865271 & 0.0865271 \
    0.45 & 0.1120350 & 0.1120346 \
    0.50 & 0.1420440 & 0.1420443 \
    0.55 & 0.1772810 & 0.1772811 \
    0.60 & 0.2187270 & 0.2187274 \
    0.65 & 0.2677510 & 0.2677515 \
    0.70 & 0.3263300 & 0.3263310 \
    0.75 & 0.3974690 & 0.3974711 \
    0.80 & 0.4860560 & 0.4860678 \
    0.85 & 0.6009030 & 0.6009565 \
    0.90 & 0.7606450 & 0.7609314 \
    0.95 & 1.0188000 & 1.0209564
    endarray
    right)$$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      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














    4












    4








    4





    $begingroup$

    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 $$J=int_0 ^afracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$



    $$left(
    beginarrayccc
    a & textapproximation & textexact \
    0.05 & 0.0012514 & 0.0012514 \
    0.10 & 0.0050231 & 0.0050231 \
    0.15 & 0.0113677 & 0.0113677 \
    0.20 & 0.0203761 & 0.0203761 \
    0.25 & 0.0321816 & 0.0321817 \
    0.30 & 0.0469674 & 0.0469674 \
    0.35 & 0.0649765 & 0.0649765 \
    0.40 & 0.0865271 & 0.0865271 \
    0.45 & 0.1120350 & 0.1120346 \
    0.50 & 0.1420440 & 0.1420443 \
    0.55 & 0.1772810 & 0.1772811 \
    0.60 & 0.2187270 & 0.2187274 \
    0.65 & 0.2677510 & 0.2677515 \
    0.70 & 0.3263300 & 0.3263310 \
    0.75 & 0.3974690 & 0.3974711 \
    0.80 & 0.4860560 & 0.4860678 \
    0.85 & 0.6009030 & 0.6009565 \
    0.90 & 0.7606450 & 0.7609314 \
    0.95 & 1.0188000 & 1.0209564
    endarray
    right)$$






    share|cite|improve this answer











    $endgroup$



    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 $$J=int_0 ^afracsin^-1 (x)(1-x^2)^frac34 ,mathrmdx$$



    $$left(
    beginarrayccc
    a & textapproximation & textexact \
    0.05 & 0.0012514 & 0.0012514 \
    0.10 & 0.0050231 & 0.0050231 \
    0.15 & 0.0113677 & 0.0113677 \
    0.20 & 0.0203761 & 0.0203761 \
    0.25 & 0.0321816 & 0.0321817 \
    0.30 & 0.0469674 & 0.0469674 \
    0.35 & 0.0649765 & 0.0649765 \
    0.40 & 0.0865271 & 0.0865271 \
    0.45 & 0.1120350 & 0.1120346 \
    0.50 & 0.1420440 & 0.1420443 \
    0.55 & 0.1772810 & 0.1772811 \
    0.60 & 0.2187270 & 0.2187274 \
    0.65 & 0.2677510 & 0.2677515 \
    0.70 & 0.3263300 & 0.3263310 \
    0.75 & 0.3974690 & 0.3974711 \
    0.80 & 0.4860560 & 0.4860678 \
    0.85 & 0.6009030 & 0.6009565 \
    0.90 & 0.7606450 & 0.7609314 \
    0.95 & 1.0188000 & 1.0209564
    endarray
    right)$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 18 at 9:47









    Peiffap

    10312




    10312










    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

















    • $begingroup$
      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












    1












    $begingroup$

    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:



    $intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x ≈ int(4x + frac383x^3) dx = frac19x^46+2x^2+C$



    $intfracsin^-1 (x)(1-x^2)^3/4 ≈ frac19x^46+2x^2+C$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      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















    1












    $begingroup$

    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:



    $intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x ≈ int(4x + frac383x^3) dx = frac19x^46+2x^2+C$



    $intfracsin^-1 (x)(1-x^2)^3/4 ≈ frac19x^46+2x^2+C$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      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













    1












    1








    1





    $begingroup$

    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:



    $intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x ≈ int(4x + frac383x^3) dx = frac19x^46+2x^2+C$



    $intfracsin^-1 (x)(1-x^2)^3/4 ≈ frac19x^46+2x^2+C$






    share|cite|improve this answer









    $endgroup$



    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:



    $intfracsin^-1 (x)(1-x^2)^3/4 ,mathrm d x ≈ int(4x + frac383x^3) dx = frac19x^46+2x^2+C$



    $intfracsin^-1 (x)(1-x^2)^3/4 ≈ frac19x^46+2x^2+C$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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
















    • $begingroup$
      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











    1












    $begingroup$

    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






    share|cite|improve this answer











    $endgroup$

















      1












      $begingroup$

      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






      share|cite|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        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






        share|cite|improve this answer











        $endgroup$



        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







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        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




        605



























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            Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye

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            Contributing propertyKeeper of the RegisterHistoric districtHistory of the National Register of Historic PlacesNational Park ServiceProperty typesCharlestonColumbiaGreenvilleNorth CharlestonRock Hill Houses on the National Register of Historic Places in South CarolinaGeorgian architecture in South CarolinaHouses completed in 1786Plantation houses in South CarolinaNational Register of Historic Places in Charleston, South CarolinaHouses in Charleston, South Carolina plantation houseAshley RiverCharlestonNational Register of Historic PlacesRevolutionary WarWilliam LowndesU.S. CongressSouth Carolina Inter-State and West Indian ExpositionGeorgianDoricbalustradespedimentoculusentablaturemodilliondentillightship roofterra cottatilescorbelledcornicesRegency stylespiral staircaseskylight Lowndes Grove U.S. National Register of Historic Places Lowndes Grove in 1940 Show map of South Carolina Show map of the United States Location 260 St. Margaret St., Charleston, South Carolina Coordinates 32...
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