Q.Integrate the following function: $\sqrt{1-4x-x^2}$

Concept: U Substitution – we rewrite the integrand as a square root of a quadratic, complete the square, and substitute to match the standard form $\sqrt{a^2 - u^2}$.

First, complete the square inside the radical:

So the integral becomes $\int \sqrt{5 - (x+2)^2} \, dx$.

Let $u = x+2$, so $du = dx$. Then we have $\int \sqrt{5 - u^2} \, du$, which is a standard form. Using the formula $\int \sqrt{a^2 - u^2} \, du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}\frac{u}{a} + C$ with $a = \sqrt{5}$: …

Why This Approach Works

When you see a square root of a quadratic like $\sqrt{ax^2 + bx + c}$, your first instinct should be: complete the square. Why? Because once the quadratic is in the form $a(x-h)^2 + k$, the expression under the root becomes something like $\sqrt{k - a(x-h)^2}$ (if $a<0$) or $\sqrt{a(x-h)^2 + k}$ (if $a>0$). These are exactly the forms that match the derivatives of inverse trigonometric functions.

Here, the quadratic is $-x^2 - 4x + 1$. The negative $x^2$ coefficient tells us we're dealing with a "backwards" parabola — so after completing the square, we'll get something like $\sqrt{5 - (x+2)^2}$. That's a perfect setup for a sine substitution: when you see $\sqrt{a^2 - u^2}$, let $u = a \sin \theta$.

Let's walk through it.

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Step-by-Step Solution

1. Complete the square inside the radical.

We have $1 - 4x - x^2$. Factor out the negative from the $x$ terms:

Now complete the square for $x^2 + 4x$. Half of 4 is 2, square it to get 4. Add and subtract 4 inside the parentheses:

Distribute the minus sign:

So the integral becomes:

2. Make a substitution to simplify the variable.

Let $u = x+2$, so $du = dx$. The integral becomes:

Now we have the classic form $\sqrt{a^2 - u^2}$ with $a = \sqrt{5}$.

3. Apply the trigonometric substitution.

For $\sqrt{a^2 - u^2}$, the standard substitution is $u = a \sin \theta$, which gives $\sqrt{a^2 - u^2} = a \cos \theta$ and $du = a \cos \theta \, d\theta$.

Let $u = \sqrt{5} \sin \theta$. Then:

• $du = \sqrt{5} \cos \theta \, d\theta$
• $\sqrt{5 - u^2} = \sqrt{5 - 5\sin^2 \theta} = \sqrt{5(1 - \sin^2 \theta)} = \sqrt{5 \cos^2 \theta} = \sqrt{5} \, |\cos \theta|$

Since we're working with a definite integral in principle (or we can restrict $\theta$ to $[-\pi/2, \pi/2]$ where $\cos \theta \ge 0$), we drop the absolute value: $\sqrt{5 - u^2} = \sqrt{5} \cos \theta$.

The integral becomes:

4. Integrate $\cos^2 \theta$.

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

Integrate term by term:

5. Convert back to $u$ (and then to $x$). …

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: Not factoring out the $-1$ before completing the square.

Why it's wrong: $1-4x-x^{2}=-(x^{2}+4x-1)=5-(x+2)^{2}$; skipping the sign flip corrupts everything. Correct approach: write it as $a^{2}-t^{2}$ with $a^{2}=5$, $t=x+2$.

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