Value of integration $\int_0^1 \sin^{-1}(3x-4x^3)dx$ I have stumbled across an definite integration problem.
The question is to evaluate $$I=\int_0^1 \sin^{-1}(3x-4x^3)dx$$
I carried out the integration as this. Let $x = \sin A$ so that A = $\sin^{-1}(x)$ then
$$\begin {align} I &=\int \sin^{-1}(3\sin A-4\sin^3A)dx\\ &=\int \sin^{-1}(\sin3A)dx\\&=\int3Adx\\&=3\int \sin^{-1}(x)\\&=3[x\sin^{-1}x+\sqrt{1-x^2}] + c\end{align}$$
Then,
$$\begin{align}I&= \biggl.3[x \sin^{-1}x+\sqrt{1-x^2}]\biggl|_0^1\\&=3[(\sin^{-1}1+\sqrt{1-1^2})-(0\times \sin^{-1}0+\sqrt{1-0})]\\&=3(\frac\pi2-1)\\&=1.71238...\end{align}$$
But the correct answer is different as given by mathematica and many other ways,
$$I=\int_0^1 \sin^{-1}(3x-4x^3)dx = 0.62535...$$
I couldn't find the mistake in my process. And I don't even know which answer is correct actually? Can you please help me here?

Note
\begin{align}
I&=\int_0^1 \sin^{-1}(3x-4x^3)dx \>\>\>\>\>\>\>(x=\sin A)\\
&= \int_0^{\frac\pi2} \sin^{-1}(\sin3A)\cos AdA \\
 &= \int_0^{\frac\pi6} \sin^{-1}(\sin3A)\cos AdA
 +  \int_{\frac\pi6}^{\frac\pi2} \sin^{-1}(\sin(\pi-3A))\cos AdA \\
 &=  \int_0^{\frac\pi6} 3A\cos AdA
 +  \int_{\frac\pi6}^{\frac\pi2} (\pi-3A)\cos AdA \\
 &=\left( \frac{3\sqrt3}2-3 +\frac\pi4 \right)+ \left(\frac{3\sqrt3}2-\frac{3\pi}4\right)\\
&=3\sqrt3-3-\frac\pi2
\end{align}
