Q.Find $\int \sqrt{3 - 2x - x^2}\, dx$

The key idea is Integration by Completing the Square — rewriting the quadratic inside the square root to match the standard form $\sqrt{a^2 - (x+b)^2}$.

First, complete the square for $3 - 2x - x^2$:

So the integral becomes:

This matches the standard form $\int \sqrt{a^2 - u^2}\, du$ with $a = 2$ and $u = x+1$. The known result is:

$\int \sqrt{a^2 - u^2}\, du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}\frac{u}{a} + C.$ …

When you see a quadratic inside a square root, the first instinct should be: can I rewrite it as something squared minus something else? That’s the heart of integration by completing the square. The expression $\sqrt{3 - 2x - x^2}$ is not a perfect square on its own, but we can force it into the form $\sqrt{a^2 - (x + b)^2}$, which matches a known trigonometric substitution.

Why does this work? Because $\sqrt{a^2 - u^2}$ is the length of a leg in a right triangle with hypotenuse $a$ and one leg $u$. That geometric link leads directly to a sine substitution ($u = a\sin\theta$) and a clean integral. The formula that emerges is a standard result — once you have it, you never need to re-derive it every time.

Let’s walk through it.

1. Complete the square inside the radical. The quadratic is $-x^2 - 2x + 3$. Factor out the negative sign from the $x^2$ and $x$ terms:

To complete the square inside the parentheses, take half of $2$ (which is $1$), square it (giving $1$), add and subtract it:

So the whole expression becomes:

Therefore:

2. Recognise the standard form. We now have $\sqrt{a^2 - u^2}$ with $a = 2$ and $u = x+1$. The integral becomes:

This is a textbook case. The formula for $\int \sqrt{a^2 - u^2}\, du$ is:

If you’ve never seen where this comes from, it’s derived by substituting $u = a\sin\theta$, then $du = a\cos\theta\, d\theta$, and $\sqrt{a^2 - u^2} = a\cos\theta$. The integral becomes $\int a^2\cos^2\theta\, d\theta$, which you solve using the double-angle identity. The result above is the cleaned-up version.

3. Apply the formula. Here $a = 2$ and $u = x+1$. So: …

Method: Complete the Square, Then Standard $\sqrt{a^2-u^2}$ Form

Use this when integrating the square root of a downward quadratic ($-x^2$ leading): recast it as $a^2-(x-h)^2$ so the arcsine-type formula applies.

Steps

Step 1: Complete the square, factoring out the $-1$.

Rewrite $3-2x-x^2=-(x^2+2x-3)=4-(x+1)^2$, so $\sqrt{3-2x-x^2}=\sqrt{2^2-(x+1)^2}$.

Step 2: Apply the standard result for $\sqrt{a^2-u^2}$.

With $u=x+1$, $a=2$, …

Common Mistakes

Mistake 1: Forgetting to factor the $-1$ from $-x^2-2x$.

Why it's wrong: without it you cannot form $a^2-(x+1)^2$, and the sign of the square goes wrong. Correct approach: pull out $-1$ before completing the square.

Mistake 2: Using the $\log$ (sum-of-squares) formula.

Why it's wrong: here the structure is $a^2-u^2$ (a difference), which gives $\sin^{-1}$, not a log. Correct approach: match the difference to the arcsine formula. …