https://file.aiquickdraw.com/mathbot/file/17278774357576aug7khh.jpg

Let's analyze and solve the given double integral step by step.

The expression in the image is:

\[
\int_0^{\frac{\pi}{2}} \int_0^{\ln(2)} y \cos(x) e^{y \sin(x)} \, dy \, dx
\]

### Step 1: Focus on the inner integral
We first solve the inner integral with respect to \( y \), treating \( x \) as a constant. The inner integral is:

\[
\int_0^{\ln(2)} y \cos(x) e^{y \sin(x)} \, dy
\]

The factor \( \cos(x) \) is constant with respect to \( y \), so it can be factored out:

\[
\cos(x) \int_0^{\ln(2)} y e^{y \sin(x)} \, dy
\]

### Step 2: Solve the inner integral
We need to solve the integral:

\[
\int_0^{\ln(2)} y e^{y \sin(x)} \, dy
\]

This is an integration by parts problem. Let’s set:

- \( u = y \), so \( du = dy \),
- \( dv = e^{y \sin(x)} \, dy \), so \( v = \frac{1}{\sin(x)} e^{y \sin(x)} \) (using the chain rule).

Using integration by parts:

\[
\int u \, dv = uv - \int v \, du
\]

Substitute the values:

\[
y \cdot \frac{1}{\sin(x)} e^{y \sin(x)} \Bigg|_0^{\ln(2)} - \int_0^{\ln(2)} \frac{1}{\sin(x)} e^{y \sin(x)} \, dy
\]

### Step 3: Simplify the result of integration by parts
Let’s compute the boundary terms:

At \( y = \ln(2) \):

\[
\frac{\ln(2)}{\sin(x)} e^{\ln(2) \sin(x)} = \frac{\ln(2)}{\sin(x)} \cdot 2^{\sin(x)}
\]

At \( y = 0 \):

\[
0 \cdot e^{0} = 0
\]

Now the integral part:

\[
- \frac{1}{\sin(x)} \int_0^{\ln(2)} e^{y \sin(x)} \, dy = - \frac{1}{\sin(x)} \cdot \frac{1}{\sin(x)} e^{y \sin(x)} \Bigg|_0^{\ln(2)}
\]

Evaluating the limits:

At \( y = \ln(2) \):

\[
\frac{1}{\sin(x)} \cdot e^{\ln(2) \sin(x)} = \frac{2^{\sin(x)}}{\sin(x)}
\]

At \( y = 0 \):

\[
\frac{1}{\sin(x)} \cdot e^0 = \frac{1}{\sin(x)}
\]

So the result of the inner integral becomes:

\[
\frac{\ln(2)}{\sin(x)} 2^{\sin(x)} - \left( \frac{2^{\sin(x)} - 1}{\sin^2(x)} \right)
\]

### Step 4: Substitute the result into the outer integral
Now we substitute this result back into the outer integral:

\[
\int_0^{\frac{\pi}{2}} \cos(x) \left( \frac{\ln(2)}{\sin(x)} 2^{\sin(x)} - \frac{2^{\sin(x)} - 1}{\sin^2(x)} \right) dx
\]

This integral looks complicated, but it can be evaluated numerically if needed.

### Step 5: Numerical evaluation
At this point, we can numerically compute the value of the outer integral using methods like numerical integration (e.g., Simpson's rule, trapezoidal rule, or software tools).

Would you like me to evaluate this integral numerically or explain more about the steps involved?

---

Here are some related questions:
1. What is the significance of integrating over different variables in this problem?
2. Can you show another example of applying integration by parts?
3. What happens if the limits of integration are changed?
4. How can we approximate such integrals using numerical methods?
5. How does the cosine factor affect the behavior of the integral?

**Tip:** When performing integration by parts, always carefully choose \( u \) and \( dv \) to simplify the resulting integrals.

Solve the given double integral: ∫₀^(π/2) ∫₀^(ln2) y cos(x) e^(y sin(x)) dy dx (x is in radians).

This problem involves solving a double integral: ∫₀^(π/2) ∫₀^(ln2) y cos(x) e^(y sin(x)) dy dx using methods like integration by parts and trigonometric identities. Suitable for students at the undergraduate calculus level or advanced high school students.

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