Evaluate $\int \frac{dt}{\sqrt{3t-2t^2}}$.

$\frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{4t - 3}{3}\right) + C$

The correct answer is Option (4) → $\frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{4t - 3}{3}\right) + C$

Let $I = \int \frac{dt}{\sqrt{3t-2t^2}} = \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{-\left(t^2 - \frac{3}{2}t\right)}}$

$= \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{-\left[\left(t^2 - 2 \times t \times \frac{3}{4} + \left(\frac{3}{4}\right)^2\right) - \left(\frac{3}{4}\right)^2\right]}}$

$= \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{-\left[\left(t - \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2\right]}}$

$= \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{\left(\frac{3}{4}\right)^2 - \left(t - \frac{3}{4}\right)^2}} = \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{t - \frac{3}{4}}{\frac{3}{4}}\right) + C$

$\Rightarrow I = \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{4t - 3}{3}\right) + C$