What method should be used to integrate $\int \sqrt{2x - x^2} \,\mathrm d x$?

Question

I've tried various different substitutions but I can't seem to be able to integrate this expression.

$$\int \sqrt{2x - x^2} \,\mathrm d x$$

I'm not looking for the whole calculation done out, just a hint would suffice. What is the next step?

Answer 1

Hint...substitute $x-1=\sin\theta$

Answer 2

$$ 2x - x^2 = 1 - (1 - x)^2 $$ If $1 - x = \sin \theta$, then $dx = -\sin \theta d\theta$ and $$ \int \sqrt{2x - x^2}dx = \int \sqrt{1 - (1 - x)^2}dx = -\int \sqrt{1 - \sin^2 \theta}\cos \theta d\theta = -\int \cos^2 \theta d\theta $$ $$ = -\dfrac{1}{2}\int [1 + \cos(2\theta)]d\theta = -\dfrac{\theta}{2} - \dfrac{1}{4}\sin(2\theta) = -\dfrac{\arcsin(1 - x)}{2} - \dfrac{(1 - x)\sqrt{2x - x^2}}{2} + C $$

Answer 3

Set $y=\sqrt{2x-x^2}$; then $y^2=2x-x^2$, so your function is part of the circle of equation $x^2+y^2-2x=0$, which has center $(1,0)$ and radius $1$. A simple translation $x=X+1$, $Y=y$ brings it in the form $X^2+Y^2=1$, so the substitution you need is $$ x-1=\cos\theta,\quad y=\sin\theta $$ for $\theta\in[0,\pi]$. Then $dx=-\sin\theta\,d\theta$ and the integral becomes $$ \int-\sin^2\theta\,d\theta $$ which is well known.

Answer 4

The integral $\,\int\sqrt{a^2-x^2}\,dx $ $\,$evaluates out to $\,\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}arcsin(\frac xa)+C$ $\,\,$(I'll leave it you to check that)

Now use completing the square on ${2x-x^2}$, which becomes $\,1-(x-1)^2$

So the integral $\,I=\int\sqrt{2x-x^2}\,dx = \int \sqrt{1-(x-1)^2}\,dx = \frac {x-1}{2}\sqrt{2x-x^2}+\frac12arcsin(\frac{x-1}{1}) + C $