## Calculus (3rd Edition)

$$\frac{1}{\sqrt 3}\sin^{-1}\left(\sqrt{\frac{3}{5}}t\right)+c.$$
Since $u=\sqrt{\frac{3}{5}}t$, then $du= \sqrt{\frac{3}{5}}dt$, and hence we have $$\int \frac{ dt}{\sqrt{5-3t^2}}=\sqrt{\frac{5}{3}}\int \frac{ du}{\sqrt{5-5u^2}}=\frac{1}{\sqrt 3}\int \frac{ du}{\sqrt{1-u^2}}\\ =\frac{1}{\sqrt 3}\sin^{-1}u+c=\frac{1}{\sqrt 3}\sin^{-1}\left(\sqrt{\frac{3}{5}}t\right)+c.$$