To evaluate the given double integral

$\int_{0}^{\sqrt{\pi}} \int_{y}^{\sqrt{\pi}} \cos(x^2) \, dx \, dy,$

we will reverse the order of integration and then solve. Here are the steps:

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Step 1: Analyze the original limits of integration

1. $y$ ranges from $0$ to $\sqrt{\pi}$.
2. For a fixed $y$, $x$ ranges from $y$ to $\sqrt{\pi}$.

Thus, the region of integration in the $xy$-plane is:

$R = \{(x, y) \mid 0 \leq y \leq \sqrt{\pi}, \, y \leq x \leq \sqrt{\pi}\}.$

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Step 2: Reverse the order of integration

To reverse the order:

• $x$ will now range from $0$ to $\sqrt{\pi}$.
• For a fixed $x$, $y$ will range from $0$ to $x$.

Thus, the new integral becomes:

$\int_{0}^{\sqrt{\pi}} \int_{0}^{x} \cos(x^2) \, dy \, dx.$

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Step 3: Simplify the inner integral

The inner integral is with respect to $y$. Since $\cos(x^2)$ is independent of $y$, we can integrate it directly:

$\int_{0}^{x} \cos(x^2) \, dy = \cos(x^2) \cdot \int_{0}^{x} 1 \, dy = \cos(x^2) \cdot x.$

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Step 4: Evaluate the outer integral

The outer integral now becomes:

$\int_{0}^{\sqrt{\pi}} x \cos(x^2) \, dx.$

Let $u = x^2$, so that $du = 2x \, dx$, or $x \, dx = \frac{1}{2} du$. The limits for $u$ are:

• When $x = 0$, $u = 0$.
• When $x = \sqrt{\pi}$, $u = \pi$.

Substituting, the integral becomes:

$\int_{0}^{\sqrt{\pi}} x \cos(x^2) \, dx = \frac{1}{2} \int_{0}^{\pi} \cos(u) \, du.$

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Step 5: Solve the integral

The integral of $\cos(u)$ is $\sin(u)$. Thus:

$\frac{1}{2} \int_{0}^{\pi} \cos(u) \, du = \frac{1}{2} \left[ \sin(u) \right]_{0}^{\pi}.$

Evaluate $\sin(u)$ at the limits:

• $\sin(\pi) = 0$,
• $\sin(0) = 0$.

Therefore:

$\frac{1}{2} \left[ \sin(\pi) - \sin(0) \right] = \frac{1}{2} (0 - 0) = 0.$

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Final Answer:

The value of the double integral is:

$\boxed{0}.$

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Would you like details on reversing integration, or any clarification? Here are 5 related questions to expand your understanding:

1. What are some general strategies for reversing the order of integration?
2. How would the solution change if the integrand was $\sin(x^2)$ instead of $\cos(x^2)$?
3. Can you visualize the region of integration for this problem?
4. Why do we use substitution when solving the outer integral?
5. How does symmetry in trigonometric functions affect the integral's result?

Tip: Always sketch the region of integration when reversing the order—it helps prevent errors.