Evaluate the following integral.
Integral from 0 to StartFraction pi Over 2 EndFraction Integral from y to StartFraction pi Over 2 EndFraction 6 sine left parenthesis 2 x minus 3 y right parenthesis dx font size decreased by 5 dy
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Part 1
Integral from 0 to StartFraction pi Over 2 EndFraction Integral from y to StartFraction pi Over 2 EndFraction 6 sine left parenthesis 2 x minus 3 y right parenthesis dx font size decreased by 5 dy
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3 ​(Type an exact​ answer, using pi as​ needed.)

We are asked to evaluate the following double integral:

\[
\int_0^{\frac{\pi}{2}} \int_y^{\frac{\pi}{2}} 6 \sin(2x - 3y) \, dx \, dy
\]

### Step 1: Evaluate the inner integral
The inner integral is with respect to \(x\), with limits from \(y\) to \(\frac{\pi}{2}\):

\[
\int_y^{\frac{\pi}{2}} 6 \sin(2x - 3y) \, dx
\]

We can perform a substitution to simplify this integral. Let:

\[
u = 2x - 3y \quad \text{so} \quad du = 2dx \quad \Rightarrow \quad dx = \frac{du}{2}
\]

When \(x = y\), \(u = 2y - 3y = -y\), and when \(x = \frac{\pi}{2}\), \(u = \pi - 3y\). Now the inner integral becomes:

\[
\int_{-y}^{\pi - 3y} 6 \sin(u) \cdot \frac{du}{2} = 3 \int_{-y}^{\pi - 3y} \sin(u) \, du
\]

### Step 2: Evaluate the sine integral
The integral of \(\sin(u)\) is \(-\cos(u)\), so we have:

\[
3 \left[ -\cos(u) \right]_{-y}^{\pi - 3y}
\]

Substitute the limits:

\[
3 \left( -\cos(\pi - 3y) + \cos(-y) \right)
\]

Using the identities \(\cos(\pi - 3y) = -\cos(3y)\) and \(\cos(-y) = \cos(y)\), this simplifies to:

\[
3 \left( -(-\cos(3y)) + \cos(y) \right) = 3 \left( \cos(3y) + \cos(y) \right)
\]

Thus, the inner integral evaluates to:

\[
3 (\cos(3y) + \cos(y))
\]

### Step 3: Evaluate the outer integral
Now, we need to compute the outer integral:

\[
\int_0^{\frac{\pi}{2}} 3 (\cos(3y) + \cos(y)) \, dy
\]

This splits into two separate integrals:

\[
3 \int_0^{\frac{\pi}{2}} \cos(3y) \, dy + 3 \int_0^{\frac{\pi}{2}} \cos(y) \, dy
\]

The integral of \(\cos(ky)\) is \(\frac{\sin(ky)}{k}\). So, we get:

\[
3 \left( \frac{\sin(3y)}{3} \right)_0^{\frac{\pi}{2}} + 3 \left( \sin(y) \right)_0^{\frac{\pi}{2}}
\]

### Step 4: Simplify the result
Evaluate both sine terms:

\[
3 \left( \frac{\sin\left( \frac{3\pi}{2} \right) - \sin(0)}{3} \right) + 3 \left( \sin\left( \frac{\pi}{2} \right) - \sin(0) \right)
\]

Since \(\sin\left( \frac{3\pi}{2} \right) = -1\) and \(\sin\left( \frac{\pi}{2} \right) = 1\), this simplifies to:

\[
3 \left( \frac{-1 - 0}{3} \right) + 3(1 - 0) = 3(-\frac{1}{3}) + 3 = -1 + 3 = 2
\]

### Final Answer:
The value of the double integral is:

\[
\boxed{2}
\]

Let me know if you'd like more details or have any further questions!

Here are 5 related questions:
1. What is the significance of using trigonometric identities in integrals?
2. How does substitution simplify double integrals?
3. What are some common techniques for solving double integrals with trigonometric functions?
4. Can double integrals be solved using numerical methods if analytic methods are difficult?
5. How does changing the order of integration affect the solution of a double integral?

**Tip:** When solving integrals with sine or cosine, knowing key trigonometric identities can simplify the process.

Evaluate the following integral: Integral from 0 to π/2, Integral from y to π/2, 6 sin(2x - 3y) dx dy.

Solve a double integral involving trigonometric functions: ∫∫ 6 sin(2x - 3y) dx dy over the range 0 to π/2. This problem demonstrates the use of substitution and trigonometric identities. A detailed solution for advanced calculus students is provided.

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