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

Concept: U Substitution (Trigonometric Substitution)

The expression $\sqrt{4-x^2}$ suggests the substitution $x = 2\sin\theta$, because $4 - x^2 = 4 - 4\sin^2\theta = 4\cos^2\theta$, and $\sqrt{4-x^2} = 2|\cos\theta|$. For $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$, $\cos\theta \ge 0$, so the absolute value drops.

Steps:

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

   $\sqrt{4-x^2} = \sqrt{4 - 4\sin^2\theta} = 2\cos\theta$.

2. The integral becomes

   $\int \sqrt{4-x^2}\,dx = \int (2\cos\theta)(2\cos\theta\,d\theta) = 4\int \cos^2\theta\,d\theta$.

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

   $4\int \frac{1+\cos2\theta}{2}\,d\theta = 2\int (1+\cos2\theta)\,d\theta = 2\theta + \sin2\theta + C$.

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

Thus the antiderivative is

$2\arcsin\left(\frac{x}{2}\right) + \frac{x\sqrt{4-x^2}}{2} + C$.

The key insight: when you see $\sqrt{a^2 - x^2}$, think of the Pythagorean identity $1 - \sin^2\theta = \cos^2\theta$. The expression under the square root is begging to become a perfect square of a cosine. That’s the heart of trigonometric substitution — it turns an algebraic square root into a clean trigonometric function.

Here, $a = 2$, so $\sqrt{4 - x^2} = \sqrt{2^2 - x^2}$. The substitution $x = 2\sin\theta$ will make $4 - x^2 = 4 - 4\sin^2\theta = 4\cos^2\theta$, and the square root becomes $2|\cos\theta|$. For the principal range $\theta \in [-\pi/2, \pi/2]$, $\cos\theta \ge 0$, so we can drop the absolute value.

Let’s work through it.

1. Set up the substitution.

   Let $x = 2\sin\theta$, so $dx = 2\cos\theta \, d\theta$.

   Then $\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = \sqrt{4(1 - \sin^2\theta)} = \sqrt{4\cos^2\theta} = 2|\cos\theta|$.

   For $\theta \in [-\pi/2, \pi/2]$, $\cos\theta \ge 0$, so $\sqrt{4 - x^2} = 2\cos\theta$.

2. Rewrite the integral.

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

This gives $2\left(\theta + \frac{1}{2}\sin 2\theta\right) + C = 2\theta + \sin 2\theta + C$.

4. Simplify $\sin 2\theta$.

   $\sin 2\theta = 2\sin\theta\cos\theta$. So the integral becomes $2\theta + 2\sin\theta\cos\theta + C$.

5. Back-substitute to $x$.

   From $x = 2\sin\theta$, we have $\sin\theta = \frac{x}{2}$.

   Then $\theta = \sin^{-1}\frac{x}{2}$.

   For $\cos\theta$, use $\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4 - x^2}}{2}$.

   Therefore:

6. Write the final antiderivative.

Method: Trigonometric Substitution for $\sqrt{a^2-x^2}$

Use this when the integrand contains $\sqrt{a^2-x^2}$: a sine substitution turns the root into a plain cosine, removing the radical.

Steps

Step 1: Match the pattern and substitute.

Recognise $\sqrt{a^2-x^2}$ and set $x=a\sin\theta$, so $dx=a\cos\theta\,d\theta$. Then

using $1-\sin^2\theta=\cos^2\theta$.

Step 2: Reduce to a power-reduction integral.

The integral becomes $\int a^2\cos^2\theta\,d\theta$; apply $\cos^2\theta=\frac{1+\cos2\theta}{2}$ to integrate.

Step 3: Back-substitute using a right triangle.

Convert $\theta$ and $\sin2\theta=2\sin\theta\cos\theta$ back to $x$ via $\sin\theta=\frac{x}{a}$ and $\cos\theta=\frac{\sqrt{a^2-x^2}}{a}$:

Common Mistakes

Mistake 1: Using $x=a\tan\theta$ for $\sqrt{a^2-x^2}$.

Why it's wrong: the tangent substitution suits $a^2+x^2$; for $a^2-x^2$ you need $x=a\sin\theta$. Correct approach: match the substitution to the sign inside the root.

Mistake 2: Forgetting $dx=a\cos\theta\,d\theta$.

Why it's wrong: omitting the $a\cos\theta$ factor drops part of the integrand. Correct approach: differentiate $x=a\sin\theta$ and substitute for $dx$ too.

Mistake 3: Failing to convert $\sin2\theta$ back to $x$.

Why it's wrong: the answer must be in $x$; leaving $\theta$ or $\sin2\theta$ is incomplete. Correct approach: use the reference triangle to rewrite everything in $x$.