 Calculate the double integral ∫∫Rxcos(2x+y)dA∫∫Rxcos⁡(2x+y)dA where RR is the region: 0≤x≤π6,0≤y≤π40≤x≤π6,0≤y≤π4



The given problem is to calculate the double integral:

\[
I = \int_{0}^{\frac{\pi}{6}} \int_{0}^{\frac{\pi}{4}} x \cos(2x + y) \, dy \, dx
\]

### Step 1: Solve the inner integral
First, let's solve the inner integral with respect to \( y \):

\[
I_1(x) = \int_{0}^{\frac{\pi}{4}} x \cos(2x + y) \, dy
\]

Since \( x \) is a constant in this inner integral, we can factor \( x \) out of the integral:

\[
I_1(x) = x \int_{0}^{\frac{\pi}{4}} \cos(2x + y) \, dy
\]

Now, the integral of \( \cos(2x + y) \) with respect to \( y \) is:

\[
\int \cos(2x + y) \, dy = \sin(2x + y)
\]

Evaluating the definite integral:

\[
I_1(x) = x \left[ \sin(2x + y) \right]_{y=0}^{y=\frac{\pi}{4}}
\]

Substitute the limits \( y = \frac{\pi}{4} \) and \( y = 0 \):

\[
I_1(x) = x \left( \sin\left(2x + \frac{\pi}{4}\right) - \sin(2x) \right)
\]

### Step 2: Solve the outer integral
Now we need to integrate \( I_1(x) \) with respect to \( x \):

\[
I = \int_{0}^{\frac{\pi}{6}} x \left( \sin\left(2x + \frac{\pi}{4}\right) - \sin(2x) \right) dx
\]

We can split this into two separate integrals:

\[
I = \int_{0}^{\frac{\pi}{6}} x \sin\left(2x + \frac{\pi}{4}\right) dx - \int_{0}^{\frac{\pi}{6}} x \sin(2x) \, dx
\]

These are standard integrals that can be solved using integration by parts.

### Step 3: Solve each integral using integration by parts

1. **First integral**:

\[
I_2 = \int_{0}^{\frac{\pi}{6}} x \sin\left(2x + \frac{\pi}{4}\right) dx
\]

Let \( u = x \), \( dv = \sin(2x + \frac{\pi}{4}) dx \).

The derivative of \( u \) is \( du = dx \), and the integral of \( dv \) is:

\[
v = -\frac{1}{2} \cos(2x + \frac{\pi}{4})
\]

Applying integration by parts:

\[
I_2 = \left[ -\frac{x}{2} \cos\left(2x + \frac{\pi}{4}\right) \right]_{0}^{\frac{\pi}{6}} + \frac{1}{2} \int_{0}^{\frac{\pi}{6}} \cos\left(2x + \frac{\pi}{4}\right) dx
\]

The second integral is straightforward:

\[
\int \cos(2x + \frac{\pi}{4}) dx = \frac{1}{2} \sin(2x + \frac{\pi}{4})
\]

Evaluating this gives:

\[
I_2 = -\frac{\pi}{12\sqrt{2}} + \frac{1}{4}
\]

2. **Second integral**:

\[
I_3 = \int_{0}^{\frac{\pi}{6}} x \sin(2x) \, dx
\]

Using integration by parts again, we get:

\[
I_3 = -\frac{\pi}{12} + \frac{1}{4}
\]

### Step 4: Final result
Now, combining the two integrals, the final result is:

\[
I = \boxed{-\frac{\pi}{12\sqrt{2}} + \frac{\pi}{12}}
\]

Would you like more detailed steps on integration by parts or clarification on a specific part?

Here are five related questions to consider:
1. How do you apply integration by parts to trigonometric functions?
2. How do you solve definite integrals involving trigonometric functions?
3. How would the result change if the region of integration were different?
4. Can you extend this to triple integrals over a similar region?
5. What are some common strategies for solving double integrals over rectangular regions?

**Tip:** Always check if the integral allows for simplifications, such as symmetry or trigonometric identities, before jumping into complex techniques like integration by parts.

Calculate the double integral ∫∫R xcos(2x+y) dA where R is the region: 0≤x≤π/6, 0≤y≤π/4

This problem demonstrates how to calculate the double integral of xcos(2x+y) over the region defined by 0≤x≤π/6 and 0≤y≤π/4. The solution involves solving trigonometric integrals and applying integration by parts, with detailed steps for each calculation. Suitable for students in Grades 11-12 or early college.

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