0
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ Complete the squares, then use trig. $\endgroup$ – lulu Mar 23 '18 at 20:33
  • 1
    $\begingroup$ Euler's substitutions solve all those problems by reducing it to integration of a rational function. $\endgroup$ – SphericalTriangle Mar 23 '18 at 20:34
  • $\begingroup$ Integrate by parts, then rework the new integrand in terms of the denominator... $\endgroup$ – Yves Daoust Mar 23 '18 at 20:36
  • 1
    $\begingroup$ Substitution is one way. Interpreting the integral as the area of part of the semi-disk and computing the area geometrically is another way. $\endgroup$ – Sangchul Lee Mar 23 '18 at 20:53
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.