Double Integral C85E11

1. **State the problem:** We need to evaluate the double integral $$\int_{x=\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_{y=\cos x}^{\frac{1}{x}} x \, dy \, dx.$$ This means integrating the function $$f(x,y) = x$$ over the region where $$x$$ ranges from $$\frac{\pi}{2}$$ to $$\frac{3\pi}{2}$$ and $$y$$ ranges from $$\cos x$$ to $$\frac{1}{x}$$. 2. **Formula and rules:** The double integral $$\int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \, dx$$ can be computed by first integrating with respect to $$y$$ treating $$x$$ as constant, then integrating the resulting expression with respect to $$x$$. 3. **Inner integral:** Since $$f(x,y) = x$$ does not depend on $$y$$, the inner integral is: $$\int_{y=\cos x}^{\frac{1}{x}} x \, dy = x \int_{\cos x}^{\frac{1}{x}} dy = x \left( \frac{1}{x} - \cos x \right) = x \cdot \frac{1}{x} - x \cos x = 1 - x \cos x.$$ 4. **Outer integral:** Now integrate with respect to $$x$$: $$\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (1 - x \cos x) \, dx = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 1 \, dx - \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} x \cos x \, dx.$$ 5. **Evaluate the first integral:** $$\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 1 \, dx = \left[ x \right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = \frac{3\pi}{2} - \frac{\pi}{2} = \pi.$$ 6. **Evaluate the second integral:** Use integration by parts for $$\int x \cos x \, dx$$. Let $$u = x$$ so $$du = dx$$. Let $$dv = \cos x \, dx$$ so $$v = \sin x$$. Then: $$\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C.$$ Evaluate from $$\frac{\pi}{2}$$ to $$\frac{3\pi}{2}$$: $$\left[ x \sin x + \cos x \right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = \left( \frac{3\pi}{2} \sin \frac{3\pi}{2} + \cos \frac{3\pi}{2} \right) - \left( \frac{\pi}{2} \sin \frac{\pi}{2} + \cos \frac{\pi}{2} \right).$$ Calculate each term: - $$\sin \frac{3\pi}{2} = -1$$ - $$\cos \frac{3\pi}{2} = 0$$ - $$\sin \frac{\pi}{2} = 1$$ - $$\cos \frac{\pi}{2} = 0$$ So: $$\left( \frac{3\pi}{2} \times (-1) + 0 \right) - \left( \frac{\pi}{2} \times 1 + 0 \right) = -\frac{3\pi}{2} - \frac{\pi}{2} = -2\pi.$$ 7. **Combine results:** The value of the outer integral is: $$\pi - (-2\pi) = \pi + 2\pi = 3\pi.$$ **Final answer:** $$\boxed{3\pi}.$$