Solve the Integral: $\int{\sqrt{8y-x^2}dx}$

Question

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?

Answer 1

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.

Answer 2

this isn't correct, Substitute $$x=\sqrt{8y}\sin(t)$$