Evaluate the integral: I = integral dx/(sqrt(4x-x²))

Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{4x-x^2}}. $$

Explanation of Method

Completing the square converts the radical into the standard $\sqrt{a^2-u^2}$ form, resolved directly by the arcsine formula — the same pattern as Problem 194 with $a=2$.

Complete the square for $4x-x^2$.
Substitute $u=x-2$.
Apply the standard formula $\int du/\sqrt{a^2-u^2}=\arcsin(u/a)$.

Step 1 · Complete the square

$$ 4x-x^2 = -\left(x^2-4x\right) = -\left[(x-2)^2-4\right] = 4-(x-2)^2. $$

Step 2 · Substitute

Let $u=x-2$, $du=dx$:

$$ I = \int \frac{du}{\sqrt{4-u^2}}. $$

Step 3 · Apply the standard arcsine formula

With $a=2$:

$$ I = \arcsin\frac u2 + C. $$

Step 4 · Substitute back

With $u=x-2$:

$$ I = \arcsin\left(\frac{x-2}{2}\right) + C. $$

Summary Table

Quantity	Value
Completed square	$4-(x-2)^2$
Substitution	$u=x-2$

Additional Notes

The domain requires $0\leq x\leq4$, where $4x-x^2\geq0$.
This is the case $a=2$ of Problem 194's general result $\int dx/\sqrt{2ax-x^2}=\arcsin\!\left(\tfrac{x-a}{a}\right)$.
The result can be checked by differentiating $\arcsin\!\left(\tfrac{x-2}{2}\right)$: this gives $\dfrac{1/2}{\sqrt{1-(x-2)^2/4}}=\dfrac{1}{\sqrt{4-(x-2)^2}}=\dfrac{1}{\sqrt{4x-x^2}}$, matching the integrand.

Result

The solution to the integral is:

$$ I = \arcsin\left(\frac{x-2}{2}\right) + C. $$