To solve the integral

$\int \sqrt{1 - x^2} \, dx,$

we can use a trigonometric substitution. The integrand suggests that we might use the substitution $x = \sin \theta$, as this simplifies the square root expression. Here's the step-by-step solution:

Step 1: Substitution

Let:

$x = \sin \theta.$

Then, the differential $dx$ is:

$dx = \cos \theta \, d\theta.$

Also, the square root expression becomes:

$\sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta.$

Step 2: Substituting into the integral

Now, we substitute into the original integral:

$\int \sqrt{1 - x^2} \, dx = \int \cos \theta \cdot \cos \theta \, d\theta = \int \cos^2 \theta \, d\theta.$

Step 3: Using a trigonometric identity

To integrate $\cos^2 \theta$, we can use the half-angle identity:

$\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}.$

Thus, the integral becomes:

$\int \cos^2 \theta \, d\theta = \int \frac{1 + \cos(2\theta)}{2} \, d\theta.$

Step 4: Integrating

Now, we can integrate term by term:

$\int \frac{1}{2} \, d\theta = \frac{\theta}{2},$

and

$\int \frac{\cos(2\theta)}{2} \, d\theta = \frac{\sin(2\theta)}{4}.$

So the integral becomes:

$\frac{\theta}{2} + \frac{\sin(2\theta)}{4} + C.$

Step 5: Back-substitution

Now, we need to back-substitute in terms of $x$. Recall that $x = \sin \theta$, so $\theta = \arcsin(x)$. Also, using the double angle identity, $\sin(2\theta) = 2\sin \theta \cos \theta = 2x \sqrt{1 - x^2}$.

Thus, the solution is:

$\frac{\arcsin(x)}{2} + \frac{x \sqrt{1 - x^2}}{2} + C.$

Final Answer:

$\int \sqrt{1 - x^2} \, dx = \frac{\arcsin(x)}{2} + \frac{x \sqrt{1 - x^2}}{2} + C.$

Would you like further clarification or to explore related concepts?

---

Further Questions:

1. How do trigonometric substitutions help with integrals involving square roots?
2. What are other examples of integrals that can be solved using trigonometric substitution?
3. Can you solve integrals involving $\sqrt{a^2 - x^2}$ using a similar method?
4. How do we approach integrals of the form $\int \sqrt{x^2 + a^2} \, dx$?
5. What are the applications of integrals like $\int \sqrt{1 - x^2} \, dx$ in physics?

Tip: When encountering an integral with $\sqrt{1 - x^2}$, try the substitution $x = \sin \theta$ to simplify the expression.