We are tasked with solving the double integral of $y \sin(xy)$ over the region $R = [0, \frac{1}{2} \pi] \times [0, 3]$. This can be written as:

$\int_0^{\frac{1}{2}\pi} \int_0^3 y \sin(xy) \, dy \, dx$

Step 1: Solve the inner integral

The inner integral is with respect to $y$:

$\int_0^3 y \sin(xy) \, dy$

We can use integration by parts for this. Let's recall the formula for integration by parts:

$\int u \, dv = uv - \int v \, du$

Choose $u = y$ and $dv = \sin(xy) \, dy$. Then, differentiate $u$ and integrate $dv$:

• $du = dy$
• To integrate $\sin(xy)$ with respect to $y$, we treat $x$ as a constant, so:

$\int \sin(xy) \, dy = -\frac{1}{x} \cos(xy)$

Now, applying the integration by parts formula:

$\int_0^3 y \sin(xy) \, dy = \left[ -y \frac{1}{x} \cos(xy) \right]_0^3 + \int_0^3 \frac{1}{x} \cos(xy) \, dy$

Evaluating the first term:

$\left[ -y \frac{1}{x} \cos(xy) \right]_0^3 = -\frac{3}{x} \cos(3x) + 0 = -\frac{3}{x} \cos(3x)$

Now, integrate the second term:

[ \int_0^3 \frac{1}{x} \cos(xy) , dy = \frac{1}{x} \left[ \frac{1}{x} \sin(xy) \right]_0^3 = \frac{1}{x^2} \left[ \sin(3x) - \sin(0) \right] = \frac{1}{x^2} \sin(3x) ]

Thus, the inner integral becomes:

$\int_0^3 y \sin(xy) \, dy = -\frac{3}{x} \cos(3x) + \frac{1}{x^2} \sin(3x)$

Step 2: Solve the outer integral

Now, we need to evaluate the outer integral:

$\int_0^{\frac{1}{2} \pi} \left( -\frac{3}{x} \cos(3x) + \frac{1}{x^2} \sin(3x) \right) dx$

We can break this into two separate integrals:

1. $-3 \int_0^{\frac{1}{2} \pi} \frac{1}{x} \cos(3x) \, dx$
2. $\int_0^{\frac{1}{2} \pi} \frac{1}{x^2} \sin(3x) \, dx$

First integral: $-3 \int_0^{\frac{1}{2} \pi} \frac{1}{x} \cos(3x) \, dx$

This is a standard integral that typically requires a special technique (such as series expansion or numerical methods). In practice, this integral does not have a simple elementary solution. It involves the cosine integral function $\text{Ci}(x)$, but for the sake of simplicity, we can either approximate or use numerical integration methods.

Second integral: $\int_0^{\frac{1}{2} \pi} \frac{1}{x^2} \sin(3x) \, dx$

Similarly, this integral involves a singularity at $x = 0$ and may not have a straightforward elementary solution. Numerical methods would be necessary to evaluate it directly.

Step 3: Numerical Evaluation (Optional)

Given the complexity of these integrals, numerical integration (using software like MATLAB, WolframAlpha, or Python) would provide an accurate solution for both integrals.

Would you like to proceed with a numerical approximation for the final result? Or would you prefer a more detailed explanation of handling these integrals analytically?