The integral in question is:

$I = \int_0^1 \sqrt{1 - x^2} \, dx$

This integral represents the area under the curve of $y = \sqrt{1 - x^2}$ from $x = 0$ to $x = 1$, which is a quarter of the unit circle centered at the origin.

Solution using trigonometric substitution:

To evaluate this integral, we'll use the substitution $x = \sin \theta$, which simplifies the square root expression.

1. Substitution: $x = \sin \theta \quad \Rightarrow \quad dx = \cos \theta \, d\theta$ When $x = 0$, $\theta = 0$; and when $x = 1$, $\theta = \frac{\pi}{2}$.

2. Rewriting the integral: $I = \int_0^{\frac{\pi}{2}} \sqrt{1 - \sin^2 \theta} \cdot \cos \theta \, d\theta$ Using the identity $\sqrt{1 - \sin^2 \theta} = \cos \theta$, we get: $I = \int_0^{\frac{\pi}{2}} \cos^2 \theta \, d\theta$

3. Simplifying the integral: Use the double-angle identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$: $I = \int_0^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta$ Now, split the integral: $I = \frac{1}{2} \int_0^{\frac{\pi}{2}} 1 \, d\theta + \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos(2\theta) \, d\theta$

4. Evaluating the integrals:

   • The first integral is straightforward: $\frac{1}{2} \int_0^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{4}$
   • The second integral is: $\frac{1}{2} \int_0^{\frac{\pi}{2}} \cos(2\theta) \, d\theta = \frac{1}{2} \cdot \left[ \frac{\sin(2\theta)}{2} \right]_0^{\frac{\pi}{2}} = 0$ since $\sin(\pi) = \sin(0) = 0$.

5. Final answer: $I = \frac{\pi}{4}$

Thus, the value of the integral is:

$\int_0^1 \sqrt{1 - x^2} \, dx = \frac{\pi}{4}$

Would you like further clarification or a deeper breakdown of any steps?

Here are some related questions to expand on this concept:

1. How would you evaluate the integral $\int_0^a \sqrt{1 - x^2} \, dx$ for any $a \in [0,1]$?
2. Can you explain the geometrical interpretation of the integral?
3. How do trigonometric substitutions work for other types of integrals?
4. What are other common integrals involving square roots and trigonometric functions?
5. How is the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$ derived and applied?

Tip: Trigonometric substitutions are very useful when dealing with integrals involving square roots of quadratic expressions.