Where did I go wrong with this integration? $$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2-3x^2}}$$
$$\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {2(1-\frac32x^2)}}$$
$$\frac 1{\sqrt 2}\int_0^{\frac {1}{\sqrt 3}} \frac {1}{\sqrt {1-\frac32x^2}}$$
$$\bigg(\frac 1{\sqrt 2}\sin^{-1}{\sqrt {\frac 32 x}}\bigg)\bigg|_0^\frac 1{\sqrt 3}$$
$$\frac 1{\sqrt 2} \times \frac {\pi}{4}$$
However, the answer is $\frac 1{\sqrt 3} \times \frac {\pi}{4}$. Where did I go wrong?

We use $f'(x)=\frac{1}{\sqrt{1-x^2}}$ for $f(x)=\sin^{-1}(x)$ and $x\in(-1,1)$. Using the chain rule gives $f'(x)=\frac{a}{\sqrt{1-(ax)^2}}$ for $f(x)=\sin^{-1}(ax)$. Hence, we have
\begin{align*}
\int_{0}^{1/\sqrt{3}}\frac{1}{\sqrt{2-3x^2}}\mathrm dx
&=\sqrt{\frac{2}{2\cdot 3}}\int_{0}^{1/\sqrt{3}}\frac{\sqrt{\frac{3}{2}}}{\sqrt{1-\left(\sqrt{\frac{3}{2}}x\right)^2}}\mathrm dx
=\sqrt{\frac{1}{3}}\left[\sin^{-1}\left(\sqrt{\frac{3}{2}}x\right)\right]_0^{1/\sqrt{3}}\\
&=\sqrt{\frac{1}{3}}\left(\frac{\pi}{4}-0\right)
=\frac{\pi}{4\sqrt{3}}.
\end{align*}
