Complicated integration with $\arcsin$ problem > $\int_{1}^{-1}\sqrt {8-3x^2}dx$ I was instructed to use substitution $x=\sqrt{\frac{8}{3}}\sin v$ but I think I mess something cause I get $\cos (\arcsin x)$ and don't now what to do with that?

Let $\displaystyle\arcsin x=\phi\implies\sin\phi=x$ and $-\frac\pi2\le \phi\le\frac\pi2$ based the principal value of inverse sine ratio

So, $\displaystyle\cos(\arcsin x)=\cos\phi=+\sqrt{1-\sin^2\phi}=+\sqrt{1-x^2}$