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

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

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