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?
 A: 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*}
A: Let  $x=\frac{\sqrt{2} \sin \theta}{\sqrt{3}}$, then $dx=\frac{\sqrt{2}}{\sqrt{3}} \cos \theta d \theta$
$$
\begin{aligned} \\
I & =\int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{2-2 \sin ^2 \theta}} \frac{\sqrt{2}}{\sqrt{3}} \cos \theta d \theta \\
& =\frac{1}{\sqrt{3}} \int_0^{\frac{\pi}{4}} d \theta \\
& =\frac{\pi}{4 \sqrt{3}}
\end{aligned}
$$
A: In case you're not required to rely on trigonometric substitutions, we can make an alternative substitution of
$$t = \frac{\sqrt{2-3x^2}-\sqrt2}x \implies x=-\frac{2\sqrt2\,t}{3+t^2} \implies dx = -\frac{2\sqrt2(3-t^2)}{(3+t^2)^2} \, dt$$
Then
$$\begin{align*}
I &= \int_0^{\frac1{\sqrt3}} \frac{dx}{\sqrt{2-3x^2}} \\[1ex]
&= 2 \int_{\sqrt3-\sqrt6}^0 \frac{dt}{3+t^2}\\[1ex]
&= - \frac2{\sqrt3} \tan^{-1}\left(\frac{\sqrt3-\sqrt6}{\sqrt3}\right) \\[1ex]
&= \frac2{\sqrt3} \tan^{-1}\left(\sqrt2-1\right) = \boxed{\frac\pi{4\sqrt3}}
\end{align*}$$
