If   $\int \frac{dt}{\sqrt{3at-2t^2}}=a^x{\sin }^{-1}\left(\frac{t^2}{a^2}-1\right)$ the value of x is _________

$\left(\frac{t^2}{a^2}-1\right)$ is dimensionless $\pm t=\pm a$

$\begin{array}{l}\left[a\right]=\left[t\right]\\ \sqrt{\left(3at-2t^2\right)}=\left[t\right]\\ \left[\frac{dt}{\sqrt{3at-2t^2}}\right]=\frac{\left[t\right]}{\left[t\right]}=\left[M^0{L}^0{T}^0\right]\\ a^x\text{ should be dimensionless, so }x=0\text{ . }\end{array}$