Calculate the double integral \( \iint_{\mathbf{R}} x \cos (2 x+y) d A \) where \( \mathbf{R} \) is the region: \( 0 \leq x \leq \frac{\pi}{3}, 0 \leq y \leq \frac{\pi}{2} \)

To calculate the double integral \[ \iint_{\mathbf{R}} x \cos (2x + y) \, dA \] where the region \(\mathbf{R}\) is defined by \(0 \leq x \leq \frac{\pi}{3}\) and \(0 \leq y \leq \frac{\pi}{2}\), we can express the double integral as: \[ \int_0^{\frac{\pi}{3}} \int_0^{\frac{\pi}{2}} x \cos(2x + y) \, dy \, dx \] ### Step 1: Integrate with respect to \(y\) First, we will integrate the inner integral with respect to \(y\): \[ \int_0^{\frac{\pi}{2}} x \cos(2x + y) \, dy \] To perform this integration, we can use the fact that the integral of \(\cos\) is \(\sin\): \[ \int \cos(2x + y) \, dy = \sin(2x + y) + C \] Now, we evaluate the definite integral: \[ \int_0^{\frac{\pi}{2}} x \cos(2x + y) \, dy = x \left[ \sin(2x + y) \right]_0^{\frac{\pi}{2}} \] Calculating the limits: \[ = x \left( \sin(2x + \frac{\pi}{2}) - \sin(2x) \right) \] Using the identity \(\sin(a + \frac{\pi}{2}) = \cos(a)\): \[ = x \left( \cos(2x) - \sin(2x) \right) \] ### Step 2: Integrate with respect to \(x\) Now we need to integrate the result with respect to \(x\): \[ \int_0^{\frac{\pi}{3}} x \left( \cos(2x) - \sin(2x) \right) \, dx \] This can be split into two separate integrals: \[ \int_0^{\frac{\pi}{3}} x \cos(2x) \, dx - \int_0^{\frac{\pi}{3}} x \sin(2x) \, dx \] ### Step 3: Calculate each integral 1. **Integral of \(x \cos(2x)\)**: Using integration by parts, let \(u = x\) and \(dv = \cos(2x) \, dx\). Then \(du = dx\) and \(v = \frac{1}{2} \sin(2x)\). Applying integration by parts: \[ \int x \cos(2x) \, dx = \frac{1}{2} x \sin(2x) - \int \frac{1}{2} \sin(2x) \, dx \] The integral of \(\sin(2x)\) is \(-\frac{1}{2} \cos(2x)\): \[ = \frac{1}{2} x \sin(2x) + \frac{1}{4} \cos(2x) + C \] Evaluating from \(0\) to \(\frac{\pi}{3}\): \[ \left[ \frac{1}{2} x \sin(2x) + \frac{1}{4} \cos(2x) \right]_0^{\frac{\pi}{3}} \] Calculating at the limits: At \(x = \frac{\pi}{3}\): \[ = \frac{1}{2} \cdot \frac{\pi}{3} \cdot \sin\left(\frac{2\pi}{3}\right) + \frac{1}{4} \cos\left(\frac{2\pi}{3}\right) \] At \(x = 0\): \[ = 0 \] Now, \(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\) and \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\): \[ = \frac{1}{2} \cdot \frac{\pi}{3} \cdot \frac{\sqrt{3}}{2} + \frac{1}{4} \left(-\frac{1}{2}\right) \] Calculating this gives: \[ = \frac{\pi \sqrt{3}}{12} - \frac{1}{8} \] 2. **Integral of \(x \sin(2x)\)**: Using integration by parts again, let \(u = x\) and \(dv = \sin(2x) \, dx\). Then \(du = dx\) and \(v = -\frac{1}{2} \cos(2x)\). Applying integration by parts: \[ \int x \sin(2x) \, dx = -\frac{1}{2} x \cos(2x) + \int \frac{1}{2} \cos(2x) \, dx \] The integral of \(\cos(2x)\) is \(\frac{1}{2} \sin(2x)\): \[ = -\frac{1}{2} x \cos(2x) + \frac{1}{4} \sin(2x) + C \] Evaluating from \(0\) to \(\frac{\pi}{3}\): \[ \left[ -\frac{1}{2} x \cos(2x) + \frac{1}{4} \sin(2x) \right]_0^{\frac{\pi}{3}} \] Calculating at the limits: At \(x = \frac{\pi}{3}\): \[ = -\frac{1}{2} \cdot \frac{\pi}{3} \cdot \cos\left(\frac{2\pi}{3}\right) + \frac{1}{4} \sin\left(\frac{2\pi}{3}\right) \] At \(x = 0\): \[ = 0 \] Now, \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\) and \(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\): \[ = -\frac{1}{2} \cdot \frac{\pi}{3} \cdot \left(-\frac{1}{2}\right) + \frac{1}{4} \cdot \frac{\sqrt{3}}{2} \] Calculating this gives: \[ = \frac{\pi}{12}