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Integrate $$\int\frac{\sin^{-1} (x)}{(1-x^2)^{\frac{3}{4}}} \,\mathrm d x$$

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:

$$\int\frac{1}{\sqrt{1-x^4}}$$

Peiffap
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Sanya
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  • Welcome to MSE. Please show us the steps that you have taken. – José Carlos Santos Mar 17 '19 at 12:34
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    Did you mean to change the exponent from $3/2$ in the linked source to $3/4$? Because the consequences are horrible. – J.G. Mar 17 '19 at 12:59
  • 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. – Sanya Mar 17 '19 at 13:02
  • When you write $sin^{-1}(x)$, do you mean $arcsin(x)$ or $\frac{1}{sin(x)}$? – Jon due Mar 17 '19 at 14:32
  • @Jondue arcsin(x) or the inverse function of sinus – Sanya Mar 17 '19 at 14:40
  • @J.G. more horrible than I expected. – Subhasis Biswas Mar 17 '19 at 14:59
  • http://m.wolframalpha.com/input/?i=arcsinx%2F%281-x%5E2%29%5E%283%2F4%29+integral – Subhasis Biswas Mar 17 '19 at 15:04
  • I would first try $x=\sin(\theta)$. Oh, I thought the exponent was $\frac32$. This is more difficult, but I'd still try that substitution. – robjohn Mar 17 '19 at 21:48
  • @robjohn, the $\sin\theta$ substitution is what's done in the webpage the OP links to, where the problem does have a $3/2$ for the exponent. – Barry Cipra Mar 17 '19 at 22:25
  • @BarryCipra: yes, I know that the substitution works for the exponent of $3/2$, but I don't see that anything can be done with the exponent $3/4$. Getting rid of the $\arcsin$ might be a start. Mathematica returns an answer with a couple of hypergeometric functions, so it's not going to be pretty in any case. – robjohn Mar 17 '19 at 22:58
  • @robjohn, all I meant was the the OP had probably already tried the $\sin\theta$ substitution. Your comment sounded like you hadn't seen what was done in the link. I agree, changing $3/2$ to $3/4$ seems to make the problem a whole lot harder. – Barry Cipra Mar 17 '19 at 23:02
  • The last integral is a bona-fide elliptic integral. It is not simply that we haven't find an elementary expression for this, but it is actually proved that this cannot be written in elementary terms. – Sangchul Lee Mar 17 '19 at 23:09

3 Answers3

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If the solution in terms of hypergeometric functions $$I=\int\frac{\sin^{-1} (x)}{(1-x^2)^{\frac{3}{4}}} \,\mathrm{d}x$$ $$I=-\frac{1}{2} \sqrt[4]{1-x^2} \left(\frac{\pi \sqrt{1-x^2} \, _3F_2\left(\frac{3}{4},\frac{3}{4},1;\frac{5}{4},\frac{7}{4};1-x^2\right)}{\sqrt {2} \Gamma \left(\frac{5}{4}\right) \Gamma \left(\frac{7}{4}\right)}+4 x \, _2F_1\left(\frac{3}{4},1;\frac{5}{4};1-x^2\right) \sin ^{-1}(x)\right)$$ 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 $$x=\sin(y) \implies I=\int\frac{y}{\sqrt{\cos (y)}}\,\mathrm{d}y$$ Expanding $\frac{1}{\sqrt{\cos (y)}}$ as a Taylor series built at $y=0$,we end with $$\frac{y}{\sqrt{\cos (y)}}=y+\frac{y^3}{4}+\frac{7 y^5}{96}+\frac{139 y^7}{5760}+\frac{5473 y^9}{645120}+\frac{51103 y^{11}}{16588800}+\frac{34988647 y^{13}}{30656102400}+O\left(y^{15}\right)$$ Integrating termwise, then $$I=\frac{y^2}{2}+\frac{y^4}{16}+\frac{7 y^6}{576}+\frac{139 y^8}{46080}+\frac{5473 y^{10}}{6451200}+\frac{51103 y^{12}}{199065600}+\frac{34988647 y^{14}}{429185433600}+O\left(y^{16}\right)$$ where $y=\sin^{-1} (x)$.

Now, a few values for $$J=\int_0 ^a\frac{\sin^{-1} (x)}{(1-x^2)^{\frac{3}{4}}} \,\mathrm{d}x$$

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

Peiffap
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Claude Leibovici
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  • It might be interesting to show how you arrived at the first expression using hypergeometric functions, even if it isn't considered acceptable. – Peiffap Mar 18 '19 at 09:08
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    @Peiffap. This is the answer given by WA (see GuterBraten's answer). – Claude Leibovici Mar 18 '19 at 09:11
  • 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. – Peiffap Mar 18 '19 at 09:13
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    @Peiffap. I agree with you but .... By the way, we could have something in the middle, still arriving to hypergeometric functions. Consider that $$\frac{1}{(1-x^2)^{\frac{3}{4}}}=\sum_{p=0}^\infty \frac{\Gamma \left(p+\frac{3}{4}\right)}{\Gamma \left(\frac{3}{4}\right) \Gamma(p+1)}x^{2p} $$ making that we are left with $$\int \sin^{-1}(x), x^{2p},dx=\frac{x^{2 n+1} \left(2 (n+1) \sin ^{-1}(x)-x , _2F_1\left(\frac{1}{2},n+1;n+2;x^2\right)\right)}{2 (n+1) (2 n+1)}$$ – Claude Leibovici Mar 18 '19 at 09:30
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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:

$\int\frac{\sin^{-1} (x)}{(1-x^2)^{3/4}} \,\mathrm d x ≈ \int(4x + \frac{38}{3}x^3) dx = \frac{19x^4}{6}+2x^2+C$

$\int\frac{\sin^{-1} (x)}{(1-x^2)^{3/4}} ≈ \frac{19x^4}{6}+2x^2+C$

GuterBraten
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  • I think that there is mistake in the Taylor expansion since, if I am not mistaken, the integrand is $x+\frac{11 x^3}{12}+O\left(x^4\right)$ making the integral to be $\frac{x^2}{2}+\frac{11 x^4}{48}+O\left(x^5\right)$. Using is for an upper bounf equal to $\frac 12$, this would give $\frac{107}{768}\approx 0.139323$ which is not bad compared to the value in the table I produced. – Claude Leibovici Mar 18 '19 at 11:34
  • I might have made an error or it is bescause i only made an approximation to order 3 – GuterBraten Mar 18 '19 at 12:56
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maybe this can help you:

For clarity I'll change the notation, $\sin^{-1}(x)=\mathrm{arcsin}(x)$.

\begin{equation} \int \frac{\mathrm{arcsin}(x)}{(1-x^2)^{3/4}} dx \end{equation}

\begin{eqnarray} y=\mathrm{arcsin}(x) &\quad& x=\sin(y)\\ dy=\frac{1}{(1-x^2)^{1/2}}dx &\quad& dx= (1-x^2)^{1/2}dy \end{eqnarray}

\begin{equation} \int \frac{\mathrm{arcsin}(x)}{(1-x^2)^{3/4}} dx = \int \frac{\mathrm{arcsin}(x)}{(1-x^2)^{1/4}(1-x^2)^{1/2}} dx \end{equation}

Applying the change of variables

\begin{equation} \int \frac{y}{(1-\sin^2(y))^{1/4}} dy = \int \frac{y}{\sqrt{\cos(y)}}dy \end{equation}

eyeballfrog
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Carlos E. González C.
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