Evaluate the integral: I = integral dx/(sqrt(2ax-x²))

Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{2ax-x^2}}. $$

Step 1 · Complete the square

$$ 2ax-x^2 = -\left(x^2-2ax\right) = -\left[(x-a)^2-a^2\right] = a^2-(x-a)^2. $$

Step 2 · Substitute

Let $u=x-a$, $du=dx$:

$$ I = \int \frac{du}{\sqrt{a^2-u^2}}. $$

Step 3 · Apply the standard arcsine formula

$$ I = \arcsin\frac ua + C. $$

Step 4 · Substitute back

With $u=x-a$:

$$ I = \arcsin\left(\frac{x-a}{a}\right) + C. $$

Summary Table

Additional Notes

• The domain requires $0\leq x\leq 2a$ (for $a>0$), where $2ax-x^2\geq0$.
• This is structurally identical to Problem 196, $\int dx/\sqrt{4x-x^2}$, which is simply the case $a=2$.
• The result can be checked by differentiating $\arcsin\!\left(\tfrac{x-a}{a}\right)$: this gives $\dfrac{1/a}{\sqrt{1-(x-a)^2/a^2}}=\dfrac{1}{\sqrt{a^2-(x-a)^2}}=\dfrac{1}{\sqrt{2ax-x^2}}$, matching the integrand.