To find the integral of $\sqrt{1 - 4x^2}$ using substitution, follow these steps:

Step 1: Substitution

Let’s use a trigonometric substitution to simplify the integral. The expression inside the square root, $1 - 4x^2$, suggests a trigonometric substitution of the form: $x = \frac{1}{2} \sin(\theta)$ This is because $\sin^2(\theta)$ will simplify the square root.

Now, differentiate $x = \frac{1}{2} \sin(\theta)$ with respect to $\theta$: $dx = \frac{1}{2} \cos(\theta) d\theta$

Step 2: Rewrite the Integral

Substituting $x = \frac{1}{2} \sin(\theta)$ and $dx = \frac{1}{2} \cos(\theta) d\theta$ into the integral: $\int \sqrt{1 - 4x^2} \, dx = \int \sqrt{1 - 4\left(\frac{1}{2} \sin(\theta)\right)^2} \cdot \frac{1}{2} \cos(\theta) \, d\theta$ Simplify the expression inside the square root: $\int \sqrt{1 - \sin^2(\theta)} \cdot \frac{1}{2} \cos(\theta) \, d\theta$ Using the identity $\cos^2(\theta) = 1 - \sin^2(\theta)$, the integral becomes: $\int \cos(\theta) \cdot \frac{1}{2} \cos(\theta) \, d\theta$ $\int \frac{1}{2} \cos^2(\theta) \, d\theta$

Step 3: Simplify the Integral

Now, use the identity for $\cos^2(\theta)$: $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$ Thus, the integral becomes: $\frac{1}{2} \int \frac{1 + \cos(2\theta)}{2} \, d\theta$ $\frac{1}{4} \int (1 + \cos(2\theta)) \, d\theta$ This separates into two integrals: $\frac{1}{4} \left( \int 1 \, d\theta + \int \cos(2\theta) \, d\theta \right)$

Step 4: Solve the Integral

Integrate each part: $\frac{1}{4} \left( \theta + \frac{1}{2} \sin(2\theta) \right) + C$ Where $C$ is the constant of integration.

Step 5: Back-Substitute $\theta$

Recall that $\sin(\theta) = 2x$, so: $\theta = \arcsin(2x)$ Also, using the double-angle identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, substitute back $\sin(\theta) = 2x$ and $\cos(\theta) = \sqrt{1 - 4x^2}$: $\frac{1}{4} \left( \arcsin(2x) + x\sqrt{1 - 4x^2} \right) + C$

Thus, the integral of $\sqrt{1 - 4x^2}$ is: $\int \sqrt{1 - 4x^2} \, dx = \frac{1}{4} \left( \arcsin(2x) + x\sqrt{1 - 4x^2} \right) + C$

Would you like any more details on this process or any further explanation?

Related Questions:

1. How would the integral change if the function inside the square root was different?
2. Can the integral be evaluated using other methods, like integration by parts?
3. What is the geometric interpretation of this integral?
4. How can this integral be applied in physics or engineering contexts?
5. What is the definite integral of $\sqrt{1 - 4x^2}$ over its domain?

Tip:

When using trigonometric substitution, always carefully back-substitute the original variable to express the final answer in the same terms as the original integral.