How to integrate definite integral of form $\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx$ During WKB approximation maths I typically end up deducing integrals that need to be calculated of the form:
$$\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx$$
where a is a constant.
I know the following standard integral
$$\int_0^{1} \sqrt{1-\sqrt{y}} \space dy = \frac{8}{15}$$
My question is how do I Use the standard integral above  to deduce that
$$\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx = \frac{8 a^{\frac{5}{2}}}{15}$$

Let $x=a^2y$. Then we get that:
\begin{equation}
\begin{split}
\int_0^{a^2}\sqrt{a-\sqrt{x}}dx & =\int_0^{1}a^2\sqrt{a-\sqrt{a^2y}}dy\\
&=\int_0^{1}a^2\sqrt{a-a\sqrt{y}}dy\\
&=\int_0^1a^2\sqrt{a(1-\sqrt{y})}dy\\
&=\int_0^1a^2a^{\frac{1}{2}}\sqrt{1-\sqrt{y}}dy\\
&=a^{\frac{5}{2}}\int_0^1\sqrt{1-\sqrt{y}}dy\\
&=\frac{8a^{\frac{5}{2}}}{15}
\end{split}
\end{equation}
