Compute $\int_0^x \sqrt{s(2-s)}\, ds $ Could anyone shed some light on how to compute the following two integrals?
$$ \int_0^x \sqrt{s(2-s)}\, ds \ \ \textrm{ for $0<x<1$}, $$
and 
$$\int_x^0 \sqrt{s(s-2)}\, ds \ \ \textrm{ for $-1<x<0$}. $$

Solution without Finding Antiderivative: Observe
\begin{align}
\int^x_0\sqrt{2s-s^2}\ ds = \int^x_0\sqrt{1-(1-s)^2}\ ds = \int^{1}_{1-x} \sqrt{1-s^2}\ ds
\end{align}
which is just the area underneath the unit semi-circle from $1-x$ to $1$. But that area is given by sector of the disk minus a right triangle, i.e. we have
\begin{align}
\int^1_{1-x} \sqrt{1-s^2}\ ds =&\ \text{ area of sector} - \text{ area of triangle}\\
=&\frac{1}{2}\theta -\frac{1}{2}(1-x)\sqrt{1-(1-x)^2} \\
=&\  \frac{1}{2}\arctan \frac{\sqrt{1-(1-x)^2}}{1-x} - \frac{1}{2}(1-x)\sqrt{1-(1-x)^2}
\end{align}
where $\theta = \arctan(y/x)$ is the angle of the sector. 
Additional Remark: Notice the area underneath the unit semi-circle from $1-x$ to $1$ is also equal to the area of quarter disk in the first quadrant minus the area underneath the unit semi-circle from $0$ to $1-x$, let us call that $A$. Observe $A$ is actually the sum of a sector and a right triangle, i.e. 
\begin{align}
A =&\ \text{ right triangle } + \text{ sector of disk}\\
=&\ \frac{1}{2} (1-x)\sqrt{1-(1-x)^2} + \frac{1}{2}\left(\frac{\pi}{2}-\theta\right).
\end{align} 
Using the trig identity
\begin{align}
\sin\left(\frac{\pi}{2}-\theta \right) = \cos\theta
\end{align}
and the fact that
\begin{align}
\tan\theta = \frac{\sqrt{1-(1-x)^2}}{1-x}
\end{align}
then we arrive at the conclusion that
\begin{align}
\cos\theta = 1-x.
\end{align}
Hence it follows
\begin{align}
\frac{\pi}{2}-\theta = \arcsin\cos\theta = \arcsin(1-x)
\end{align}
which also means
\begin{align}
A = \frac{1}{2}(1-x)\sqrt{1-(1-x)^2} + \frac{1}{2}\arcsin(1-x)
\end{align}
and
\begin{align}
\int^1_{1-x}\sqrt{1-s^2}\ ds =&\ \frac{\pi}{4}- A\\
=&\ \frac{\pi}{4}-\frac{1}{2}(1-x)\sqrt{1-(1-x)^2} - \frac{1}{2}\arcsin(1-x).
\end{align}
