<s>[INST] Solve the Integral: $\int{\sqrt{8y-x^2}dx}$ I am trying to solve the following integral:
$$\int{\sqrt{8y-x^2}}dx$$
For which I don't understand clearly how to work with root values.
What I've tried:
$$
\int{\sqrt{8y-x^2}dx} = \int{(8y-x^2)^{\frac{1}{2}}dx} \\
= \frac{(8y-x^2)^{\frac{3}{2}}}{\frac{3}{2}}\int{(8y-x^2)dx} \\
= \frac{(8y-x^2)^{\frac{3}{2}}}{\frac{3}{2}}(-\frac{x^3}{3}) \\
= -\frac{2}{3}(8y-x^2)^{\frac{3}{2}}(\frac{x^3}{3}) \\
= -\frac{2}{9}(8y-x^2)^{\frac{3}{2}}x^3
$$
Is that correct?
 [/INST] Note that if we use the substitution, $x=\sqrt{8y}\sin u$, then the integrand can be easily reduced to: 
$$I=\int \sqrt{8y-x^2} dx = \sqrt{8y} \int \cos u \sqrt{8y-8y\sin^2 u} \,du=8y \int \cos^2 u \,du$$
I hope you can take it from here.
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