How do you integrate $\int \frac{1}{\sqrt{2ax-x^2}}dx$?

Can you help me with this integral? $$\int \frac{1}{\sqrt{2ax-x^2}}dx$$ I know that the denominator can be rewritten as $$\sqrt{a^2-(x-a)^2}$$. But what should I do next. I know the that the result must contain $$\arcsin(x)$$.

• Like this – logarithm May 15 at 22:07
• You can also solve it like this, but anyone can change the problem just a little bit and you won't be able to solve it anymore. – logarithm May 15 at 22:09
• Start by substituting $u=\frac{x-a}{a}.$ Then use a trig substitution for $u$. – Robert Shore May 15 at 22:10
• Integrating $1/\sqrt(a^2-y^2)$ w.r.t $y$ is standard. There is a formula first. Then, eventually setting $y=au$ we can assume $a=1$. Now one can use the substitution $y = \sin s$, so $\sqrt{1-y^2}$ gets rid of the radical... – dan_fulea May 15 at 22:12

Actually, that integral is equal to$$\int\frac1{\sqrt{a^2-(x-a)^2}}\,\mathrm dx=\frac1a\int\frac1{\sqrt{1-\left(\frac{x-a}a\right)^2}}\mathrm dx=\arcsin\left(\frac{x-a}a\right)+c.$$
• Why in the final result there is no $\dfrac{1}{a}$. You factor $\dfrac{1}{a}$ out of the integral sign. Shouldn't it be $\dfrac{1}{a}\arcsin(\dfrac{x-a}{a})+C$ – James Warthington May 15 at 22:40
• No, because when I derive $\arcsin\left(\frac{x-a}a\right)$ I get that $\frac1a$. Or, if you wish, since$$\frac1a\int\frac1{\sqrt{1-\left(\frac{x-a}a\right)^2}}\mathrm dx=\int\frac{\frac1a}{\sqrt{1-\left(\frac{x-a}a\right)^2}}\mathrm dx,$$you can do the substitution $\frac{x-a}a=y$ and $\frac{\mathrm dx}a=\mathrm dy$, thereby getting $\int\frac1{\sqrt{1-y^2}}\,\mathrm dy=\arcsin(y)=\arcsin\left(\frac{x-a}a\right)$. – José Carlos Santos May 15 at 22:44
We have: $$\int \frac{1}{\sqrt{a^2-(x-a)^2}}dx ={1\over a} \int \frac{1}{\sqrt{1-(x/a-1)^2}}dx=$$ and write $$t= x/a-1$$ $$= \int \frac{1}{\sqrt{1-t^2}}dt=\arcsin ({x\over a}-1)+c$$
$$\int{\frac{1}{\sqrt{2ax-{{x}^{2}}}}}dx=\int{\frac{1}{\sqrt{{{a}^{2}}-{{a}^{2}}+2ax-{{x}^{2}}}}}dx=\int{\frac{1}{\sqrt{{{a}^{2}}-{{(x-a)}^{2}}}}dx=arc\sin }\frac{x-a}{a}$$