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?

Let  $x=\frac{\sqrt{2} \sin \theta}{\sqrt{3}}$, then $dx=\frac{\sqrt{2}}{\sqrt{3}} \cos \theta d \theta$
$$
\begin{aligned} \\
I & =\int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{2-2 \sin ^2 \theta}} \frac{\sqrt{2}}{\sqrt{3}} \cos \theta d \theta \\
& =\frac{1}{\sqrt{3}} \int_0^{\frac{\pi}{4}} d \theta \\
& =\frac{\pi}{4 \sqrt{3}}
\end{aligned}
$$
