Q.Integrate the following function: $\frac{1}{\sqrt{8 + 3x - x^2}}$

Concept: U Substitution — we complete the square inside the square root to match the form $\frac{1}{\sqrt{a^2 - u^2}}$, which integrates to $\arcsin\left(\frac{u}{a}\right)$.

Step 1: Complete the square

$8 + 3x - x^2 = -(x^2 - 3x - 8) = -\left[(x - \frac{3}{2})^2 - \frac{9}{4} - 8\right] = \frac{41}{4} - (x - \frac{3}{2})^2$.

Step 2: Set $u = x - \frac{3}{2}$, so $du = dx$.

The integral becomes

$\int \frac{dx}{\sqrt{\frac{41}{4} - u^2}} = \int \frac{du}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2 - u^2}}$.

Step 3: Integrate using the standard form …

Why This Approach Works

When you see a quadratic inside a square root in the denominator — especially one that doesn't factor nicely — your first instinct should be to complete the square. Why? Because the expression $\frac{1}{\sqrt{a^2 - u^2}}$ integrates directly to $\sin^{-1}(u/a) + C$. That's the target form we're aiming for.

The quadratic here is $-x^2 + 3x + 8$. It's not a perfect square, and it's not in the form $a^2 - u^2$ yet. But with a little algebra, we can force it there.

Step-by-Step Solution

1. Complete the square on the quadratic.

Start with $8 + 3x - x^2$. Factor out the negative sign from the $x^2$ and $x$ terms:

Now complete the square inside the parentheses. Take half of $-3$, square it: $\left(\frac{-3}{2}\right)^2 = \frac{9}{4}$. Add and subtract this inside:

Distribute the negative:

So the integral becomes:

2. Factor out the constant to match the standard form.

Notice $\frac{41}{4} = \left(\frac{\sqrt{41}}{2}\right)^2$. So we have:

This is exactly $\int \frac{du}{\sqrt{a^2 - u^2}}$ with $a = \frac{\sqrt{41}}{2}$ and $u = x - \frac{3}{2}$.

3. Apply the standard formula.

The formula is: …

Method: $1/\sqrt{\text{quadratic}}$ with a negative $x^2$ term $\to \sin^{-1}$

When the quadratic under the root has a negative coefficient of $x^2$, completing the square produces the form $a^2 - u^2$, which integrates to an inverse sine.

Steps

Step 1: Factor out the $-1$ from the $x^2$ and $x$ terms.

For $c + bx - x^2$, write it as $-(x^2 - bx) + c$ before completing the square — this keeps the sign bookkeeping clean.

Step 2: Complete the square and combine constants to reach $k^2 - (x+h)^2$. …

Common Mistakes

Mistake 1: Forgetting to factor out the $-1$ before completing the square.

Why it's wrong: completing the square on $x^2 - bx - c$ instead of $-(x^2-bx)+c$ flips the sign of the constant and gives a $u^2-a^2$ (log) form instead of the correct $a^2-u^2$ (arcsin) form. Correct approach: pull the $-1$ out first so the $x^2$ coefficient inside is $+1$.

Mistake 2: Using $\log$ instead of $\sin^{-1}$. …