$\int \arcsin(3x - 4x^3)dx$ 
$$\int \arcsin(3x - 4x^3)dx$$

I think the best idea here is to use integration by parts and remove the arcsine from the integral.
$$\int \arcsin(3x - 4x^3)dx = x\arcsin(3x - 4x^4) - \int {3x(1-4x^2) \over \sqrt{1- (3x - 4x^3)^2}} dx$$
Now to integrate $\displaystyle \int {3x(1-4x^2) \over \sqrt{1- (3x - 4x^3)^2}} dx$ I tried substituting $u = 3x - 4x^3$, but then I get $\displaystyle \int {x\over \sqrt{1- u^2}} du$.
Now I need to express $x$ in terms of $u$ that is I need to solve $4x^3 - 3x + u = 0$, for which I get a hopeless answer.
Any ideas how to solve this integrand ?
 A: HINT
Use the triple angle formula.
A: HINT:
Let $\arcsin x=y\implies u=\arcsin(3x-4x^3)=\arcsin(\sin3y)$
$\implies u=n\pi+(-1)^n3y$  where $n$ is any integer.
Using Principal values, $-\dfrac\pi2\le u,y\le\dfrac\pi2$
For $n=0,u=3y\implies-\dfrac\pi2\le3y\le\dfrac\pi2\iff-\dfrac\pi6\le y\le\dfrac\pi6$
For $n=1,u=\pi-3y\implies-\dfrac\pi2\le\pi-3y\le\dfrac\pi2\iff\dfrac\pi6\le y\le\dfrac\pi2$
For $n=-1,u=-\pi-3y\implies-\dfrac\pi2\le-\pi-3y\le\dfrac\pi2\iff-\dfrac\pi2\le y\le-\dfrac\pi6$
A: Another approach:
Let, $x=\sin t\implies dx=\cos t\ dt$
\begin{align*}
\int\arcsin\left(3x-4x^3\right)\ dx&=\int\arcsin\left(3\sin t-4\sin^3t\right)\cos t\ dt\\
&=\int\arcsin(\sin3t)\cos t\ dt\\
&=\int3t\cos t\ dt\\
&=3\left[t\int\cos t\ dt-\int\left\{\dfrac{d}{dt}(t)\cdot \left(\int\cos t\ dt\right)\right\}\right]\\
&=3\left[t\cdot \sin t+\cos t\right]+c\\
&=3[\arcsin x\cdot x+\cos(\arcsin x)]+c\\
&=3[x\cdot\arcsin x+\cos(\arcsin x)]+c\\
&=3x\arcsin x+3\cos\left[\arccos\sqrt{1-x^2}\right]+c\\
&=3x\arcsin x+3\sqrt{1-x^2}+c.
\end{align*}
or, you can use  direct inverse formula: $\arcsin\left(3x-4x^3\right)=3\arcsin x$.
