If the interval is extended to $[0, \pi]$, we need to find the area of the region enclosed by the curves $y = \sin(x)$ and $y = \cos(x)$ over this interval.

Step 1: Determine the Points of Intersection

As before, $\sin(x) = \cos(x)$ at $x = \frac{\pi}{4}$. Over the interval $[0, \pi]$, the curves intersect again at $x = \frac{5\pi}{4}$, but we focus only on the region within $[0, \pi]$.

Step 2: Set Up the Integral

We now split the interval into two parts:

1. From $x = 0$ to $x = \frac{\pi}{4}$, where $\cos(x)$ is above $\sin(x)$.
2. From $x = \frac{\pi}{4}$ to $x = \frac{\pi}{2}$, where $\sin(x)$ is above $\cos(x)$.
3. From $x = \frac{\pi}{2}$ to $x = \pi$, where $\sin(x)$ is above $\cos(x)$.

Step 3: Compute the Integrals

For $x \in [0, \frac{\pi}{4}]$:

$A_1 = \int_{0}^{\frac{\pi}{4}} (\cos(x) - \sin(x)) \, dx = \sqrt{2} - 1$

For $x \in [\frac{\pi}{4}, \frac{\pi}{2}]$:

$A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin(x) - \cos(x)) \, dx = -1 + \sqrt{2}$

For $x \in [\frac{\pi}{2}, \pi]$:

Here, $\sin(x)$ is always greater than $\cos(x)$. $A_3 = \int_{\frac{\pi}{2}}^{\pi} (\sin(x) - \cos(x)) \, dx$ Let's compute $A_3$: [ A_3 = \left[-\cos(x) - \sin(x)\right]_{\frac{\pi}{2}}^{\pi} = \left[-\cos(\pi) - \sin(\pi)\right] - \left[-\cos\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right)\right] ] [ = \left[1 - 0\right] - \left[0 - 1\right] = 1 + 1 = 2 ]

Step 4: Add the Three Areas Together

$A = A_1 + A_2 + A_3 = (\sqrt{2} - 1) + (-1 + \sqrt{2}) + 2 = 2\sqrt{2} - 2 + 2 = 2\sqrt{2}$

Thus, the area of the region enclosed by the curves $y = \sin(x)$ and $y = \cos(x)$ on the interval $[0, \pi]$ is:

$\boxed{2\sqrt{2}}$

This result shows that the area is larger when the interval is extended to $[0, \pi]$ compared to $[0, \frac{\pi}{2}]$.

Do you have any questions, or would you like further details?

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Here are 5 related questions to explore:

1. How would the area change if the interval was $[0, 2\pi]$?
2. Can you find the area between $y = \sin(2x)$ and $y = \cos(2x)$ on $[0, \pi]$?
3. What happens if we only consider the interval $[\frac{\pi}{2}, \pi]$?
4. How do the areas compare if we consider $y = \sin(x)$ and $y = -\cos(x)$ on $[0, \pi]$?
5. What if the curves were $y = \sin(x)$ and $y = \sin(x - \frac{\pi}{4})$ on $[0, \pi]$?

Tip: Extending the interval often requires recalculating areas based on the changing positions of the curves relative to each other. Always check which curve is above the other within each subinterval.