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

Concept: U Substitution — we complete the square inside the square root to match the form $\sqrt{a^2 - u^2}$, then use the standard integral $\int \sqrt{a^2 - u^2}\,du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}\frac{u}{a} + C$.

Step 1: Complete the square

$1 + 3x - x^2 = -(x^2 - 3x - 1) = -\left[(x - \frac{3}{2})^2 - \frac{13}{4}\right] = \frac{13}{4} - (x - \frac{3}{2})^2$.

Step 2: Substitute

Let $u = x - \frac{3}{2}$, so $du = dx$ and $a^2 = \frac{13}{4}$, i.e. $a = \frac{\sqrt{13}}{2}$.

The integral becomes $\int \sqrt{a^2 - u^2}\,du$.

Step 3: Apply the standard formula

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

Step 4: Back-substitute

Replace $u = x - \frac{3}{2}$ and $a^2 = \frac{13}{4}$: …

Why this approach works

When you see a square root of a quadratic like $\sqrt{ax^2 + bx + c}$, the first instinct is to complete the square. That turns the expression into something like $\sqrt{A - (x - h)^2}$ or $\sqrt{(x - h)^2 + A}$, which then screams for a trigonometric substitution. Here, the coefficient of $x^2$ is negative, so we’ll get a form $\sqrt{A - (x - h)^2}$ — perfect for a sine substitution.

The goal is to eliminate the square root by using the identity $1 - \sin^2 \theta = \cos^2 \theta$. That turns the integral into something purely trigonometric, which we can handle with standard formulas.

Step-by-step solution

1. Complete the square

We have $1 + 3x - x^2$. Write it as $-(x^2 - 3x) + 1$. Complete the square inside the parentheses:

$x^2 - 3x = \left(x - \frac32\right)^2 - \frac94$

So:

$1 + 3x - x^2 = -\left[\left(x - \frac32\right)^2 - \frac94\right] + 1 = -\left(x - \frac32\right)^2 + \frac94 + 1 = \frac{13}{4} - \left(x - \frac32\right)^2$

Thus:

$\sqrt{1+3x-x^2} = \sqrt{\frac{13}{4} - \left(x - \frac32\right)^2}$

2. Choose the substitution

The expression is now $\sqrt{A^2 - u^2}$ with $A = \frac{\sqrt{13}}{2}$ and $u = x - \frac32$. The standard substitution is $u = A \sin \theta$, i.e.:

$x - \frac32 = \frac{\sqrt{13}}{2} \sin \theta$

Then $dx = \frac{\sqrt{13}}{2} \cos \theta \, d\theta$.

3. Simplify the square root

$\sqrt{\frac{13}{4} - \left(\frac{\sqrt{13}}{2} \sin \theta\right)^2} = \sqrt{\frac{13}{4} - \frac{13}{4} \sin^2 \theta} = \frac{\sqrt{13}}{2} \sqrt{1 - \sin^2 \theta} = \frac{\sqrt{13}}{2} |\cos \theta|$

We can take $\cos \theta \ge 0$ by choosing $\theta \in [-\pi/2, \pi/2]$, so $|\cos \theta| = \cos \theta$.

4. Set up the integral

The integral becomes:

$\int \sqrt{1+3x-x^2} \, dx = \int \frac{\sqrt{13}}{2} \cos \theta \cdot \frac{\sqrt{13}}{2} \cos \theta \, d\theta = \frac{13}{4} \int \cos^2 \theta \, d\theta$

5. Integrate $\cos^2 \theta$

Use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$:

$\frac{13}{4} \int \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{13}{8} \int (1 + \cos 2\theta) \, d\theta = \frac{13}{8} \left( \theta + \frac12 \sin 2\theta \right) + C$

Simplify: $\frac{13}{8} \theta + \frac{13}{16} \sin 2\theta + C$.

6. Back-substitute to $x$

We have $\sin \theta = \frac{2x - 3}{\sqrt{13}}$, so $\theta = \arcsin\!\left( \frac{2x - 3}{\sqrt{13}} \right)$. …

Method: Complete the square, then use a standard $\sqrt{\ }$ formula

To integrate $\sqrt{\text{quadratic}}$, rewrite the quadratic as $(x+p)^{2}\pm a^{2}$ or $a^{2}-(x+p)^{2}$ by completing the square, substitute $t=x+p$, and quote the matching standard integral.

Steps

Step 1: Complete the square on the quadratic under the root, so it becomes $(x+p)^{2}+k$ for some constant $k$.

Step 2: Substitute $t=x+p$ (so $dt=dx$); the integral becomes $\int\sqrt{t^{2}\pm a^{2}}\,dt$ or $\int\sqrt{a^{2}-t^{2}}\,dt$.

Step 3: Apply the correct standard formula.

$\int\sqrt{t^{2}-a^{2}}\,dt=\frac{t}{2}\sqrt{t^{2}-a^{2}}-\frac{a^{2}}{2}\log\big|t+\sqrt{t^{2}-a^{2}}\big|+C,$ …

Common Mistakes

Mistake 1: Mishandling the fractional shift when completing the square.

Why it's wrong: $1+3x-x^{2}=-\left[(x-\tfrac32)^{2}-\tfrac{13}{4}\right]=\tfrac{13}{4}-(x-\tfrac32)^{2}$; the half-integer $\tfrac32$ trips students up. Correct approach: use $(b/2)=\tfrac32$ carefully.

Mistake 2: Choosing the wrong template (log vs arcsin). …