Q.Evaluate: $\int \sqrt{2ax-x^2}\,dx$

The key idea is to rewrite the quadratic under the square root by completing the square, then use a standard trigonometric substitution.

First, complete the square inside the radical:

Thus the integral becomes

Let $u = x - a$, so $du = dx$. The integral is now the standard form:

$\int \sqrt{a^2 - u^2}\,du.$ …

Why Completing the Square Works Here

When you see a quadratic inside a square root, your first instinct might be to try a $u$-substitution. But $2ax - x^2$ is not a perfect square — it's a downward-opening parabola. The trick is to rewrite it so it looks like something you already know: $a^2 - (x-a)^2$. Why $a^2$? Because the maximum value of $2ax - x^2$ occurs at $x = a$, and that maximum is $a^2$.

Once you have $a^2 - (x-a)^2$, the expression inside the square root is exactly the form that suggests a sine substitution: $\sqrt{a^2 - u^2}$ where $u = x-a$. This is a classic pattern — the integral of $\sqrt{a^2 - u^2}$ is a standard result, and we can either derive it from scratch or use a known formula.

Let's walk through the derivation so you see why this formula works, not just that it exists.

Step-by-Step Solution

1. Complete the square inside the radical.

Start with $2ax - x^2$. Factor out a negative sign to make completing the square cleaner:

Now complete the square inside the parentheses: $x^2 - 2ax = (x-a)^2 - a^2$. So

Thus the integral becomes

2. Substitute to simplify the variable.

Let $u = x - a$, so $du = dx$. The integral is now

This is exactly the standard form. The domain of the integrand requires $|u| \le a$, which matches the original quadratic being non-negative.

3. Use a trigonometric substitution.

For $\sqrt{a^2 - u^2}$, the natural substitution is $u = a\sin\theta$, where $-\pi/2 \le \theta \le \pi/2$ (this ensures $\cos\theta \ge 0$, so we can drop absolute values). Then $du = a\cos\theta\,d\theta$, and

The integral becomes

4. Integrate $\cos^2\theta$ using the double-angle identity.

Recall $\cos^2\theta = \frac{1+\cos 2\theta}{2}$. So

Now $\sin 2\theta = 2\sin\theta\cos\theta$, so

5. Back-substitute to $u$ and then to $x$. …

Method: Integrating $\sqrt{2ax-x^2}$ by completing the square (arcsine template)

Use this for $\int\sqrt{\text{const}\cdot x-x^2}\,dx$. Completing the square gives $\sqrt{a^2-u^2}$, whose integral is the arcsine template.

Steps

Step 1: Complete the square.

so set $u=x-a$ with $\sqrt{a^2-u^2}$.

Step 2: Recall the standard integral.

Step 3: Substitute back $u=x-a$.

Replace $u$ everywhere and note $\sqrt{a^2-u^2}=\sqrt{2ax-x^2}$. …

Common Mistakes

Mistake 1: Sign error completing the square.

Why it's wrong: $2ax-x^2=a^2-(x-a)^2$; forgetting the leading $-$ on $x^2$ gives $(x-a)^2-a^2$ and the wrong template. Correct approach: factor $-1$ from $x^2-2ax$ first.

Mistake 2: Using the $\sqrt{u^2+a^2}$ (log) template. …