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

The key idea is U Substitution using a trigonometric substitution, specifically $x = \frac{1}{2}\sin\theta$, to simplify the square root.

Step 1: Let $x = \frac{1}{2}\sin\theta$, so $dx = \frac{1}{2}\cos\theta\,d\theta$. Then $1 - 4x^2 = 1 - \sin^2\theta = \cos^2\theta$, and $\sqrt{1-4x^2} = |\cos\theta|$. For the principal branch, take $\cos\theta \ge 0$.

Step 2: Substitute into the integral:

Step 3: Use $\cos^2\theta = \frac{1+\cos2\theta}{2}$:

Step 4: Back-substitute: $\theta = \arcsin(2x)$, and $\sin2\theta = 2\sin\theta\cos\theta = 2(2x)\sqrt{1-4x^2} = 4x\sqrt{1-4x^2}$. Thus:

Why U Substitution (and a Trigonometric One) Works

When you see $\sqrt{1 - 4x^2}$, your first instinct might be to try a simple $u = 1 - 4x^2$. That would give $du = -8x \, dx$, but there’s no $x$ outside the square root to pair with it — so that path dead-ends.

The deeper structure here is $\sqrt{1 - (2x)^2}$. That’s a perfect match for the Pythagorean identity: $1 - \sin^2 \theta = \cos^2 \theta$. If we set $2x = \sin \theta$, the square root becomes $\sqrt{1 - \sin^2 \theta} = |\cos \theta|$, and for the principal range we can take $\cos \theta \ge 0$. This substitution turns an algebraic mess into a clean trigonometric integral.

Step-by-Step Solution

1. Set up the substitution.

Let $2x = \sin \theta$. Then $x = \frac{1}{2} \sin \theta$, so $dx = \frac{1}{2} \cos \theta \, d\theta$.

2. Rewrite the integrand.

The square root becomes:

We restrict $\theta$ to $[-\pi/2, \pi/2]$ so that $\cos \theta \ge 0$, and we can drop the absolute value: $\sqrt{1 - 4x^2} = \cos \theta$.

3. Transform the integral.

The original integral is $\int \sqrt{1 - 4x^2} \, dx$. Substituting everything:

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

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

Then:

Integrate term by term:

5. Convert back to $x$.

We have $\theta = \sin^{-1}(2x)$. For $\sin 2\theta$, use $\sin 2\theta = 2 \sin \theta \cos \theta$.

We know $\sin \theta = 2x$ and $\cos \theta = \sqrt{1 - 4x^2}$ (from step 2).

So $\sin 2\theta = 2 \cdot (2x) \cdot \sqrt{1 - 4x^2} = 4x \sqrt{1 - 4x^2}$.

Thus:

Simplify:

Method: Scale to the standard $\int\sqrt{a^{2}-u^{2}}$ form

For $\sqrt{1-k^{2}x^{2}}$ (a difference of squares with a coefficient on $x^{2}$), substitute $u=kx$ to reach the textbook $\sqrt{a^{2}-u^{2}}$ integral.

Steps

Step 1: Identify the difference-of-squares form. $\sqrt{1-4x^{2}}=\sqrt{1-(2x)^{2}}$, so $u=2x$, $a=1$.

Step 2: Substitute $u=2x$, $du=2\,dx$, i.e. $dx=\dfrac{du}{2}$:

Step 3: Apply the standard result.

Step 4: Back-substitute $u=2x$ and multiply by $\tfrac12$.

Simplify and add $C$; the arcsine appears because the form is $a^{2}-u^{2}$.

Common Mistakes

Mistake 1: Forgetting the $\dfrac12$ from $du=2\,dx$.

Why it's wrong: without it the whole answer is doubled. Correct approach: $dx=\dfrac{du}{2}$ scales the integral by $\tfrac12$.

Mistake 2: Writing $\sqrt{1-4x^{2}}=\sqrt{1}-\sqrt{4x^{2}}$.

Why it's wrong: the root of a difference is not the difference of roots. Correct approach: keep it as $\sqrt{1-(2x)^{2}}$ and substitute.

Mistake 3: Using a logarithm instead of $\sin^{-1}$.

Why it's wrong: a $1-u^{2}$ (i.e. $a^{2}-u^{2}$) form gives arcsine. Correct approach: the answer has $\dfrac14\sin^{-1}(2x)$, not a log.