Compute $\int_0^x \sqrt{s(2-s)}\, ds $

Question

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 $$\int_x^0 \sqrt{s(s-2)}\, ds \ \ \textrm{ for $-1<x<0$}. $$

Answer 1

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}

Answer 2

$$s(2-s)=1^2-(s-1)^2$$

Now use $\#8$ of this

See also: Trouble solving $\int\sqrt{1-x^2} \, dx$

$$s(s-2)=(s-1)^2-1^2$$

Now use $\#8$ of this