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
$$\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx = \frac{8 a^{\frac{5}{2}}}{15}$$
 A: 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}
A: Hint
Let $\sqrt{a-\sqrt x}=y$
$x=(a-y)^2,dx=-2(a-y)dy$
So, we have $$-2\int_{\sqrt a}^0y(a-y)\ dy$$
Can you take it home from here?
A: Letting $\sqrt   x=a \sin^2 \theta$ transforms the integral into
$$
\begin{aligned}
I &=4\int_0^{\frac{\pi}{2}} \sqrt{a-a \sin ^2 \theta} a^2 \sin ^3 \theta \cos \theta d \theta \\
&=4 a^{\frac{5}{2}}\int_0^{\frac{\pi}{2}} \sin ^3 \theta \cos ^2 \theta d \theta \\
&=-4 a^{\frac{5}{2}}\int_0^{\frac{\pi}{2}}\left(1-\cos ^2 \theta\right) \cos ^2 \theta d(\cos \theta) \\
&=-4 a^{\frac{5}{2}}\left[\frac{\cos ^2 \theta}{3}-\frac{\cos ^5 \theta}{5}\right]_0^{\frac{\pi}{2}} \\
&=\frac{8 a^{\frac{5}{2}}}{15}
\end{aligned}
$$
