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

Question

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:

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

Answer 1

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

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

Answer 2

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:

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

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

Answer 3

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}