We aim to evaluate the double integral:

$\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} 2x \sin(xy) \, dy \, dx.$

Step 1: Integrate with respect to $y$

Fix $x$ as a constant and integrate $2x \sin(xy)$ with respect to $y$. The integral of $\sin(xy)$ is:

$\int \sin(xy) \, dy = -\frac{\cos(xy)}{x}.$

So, the inner integral becomes:

$\int_{0}^{1} 2x \sin(xy) \, dy = 2x \left[-\frac{\cos(xy)}{x}\right]_{y=0}^{y=1}.$

Simplify the expression:

[ = 2x \left[-\frac{\cos(x \cdot 1)}{x} + \frac{\cos(x \cdot 0)}{x}\right] = 2x \left[-\frac{\cos(x)}{x} + \frac{\cos(0)}{x}\right]. ]

Since $\cos(0) = 1$, this becomes:

= 2x \cdot \frac{-\cos(x) + 1}{x}.$$ Simplify further: $$= 2(-\cos(x) + 1) = 2 - 2\cos(x).$$ So the inner integral evaluates to: $$\int_{0}^{1} 2x \sin(xy) \, dy = 2 - 2\cos(x).$$ --- ### Step 2: Integrate with respect to $$x$$ Now integrate the result with respect to $$x$$: $$\int_{0}^{\frac{\pi}{2}} (2 - 2\cos(x)) \, dx.$$ Split the integral: $$\int_{0}^{\frac{\pi}{2}} 2 \, dx - \int_{0}^{\frac{\pi}{2}} 2\cos(x) \, dx.$$ 1. **First term**: $$\int_{0}^{\frac{\pi}{2}} 2 \, dx = 2x \bigg|_{0}^{\frac{\pi}{2}} = 2\left(\frac{\pi}{2}\right) - 2(0) = \pi.$$ 2. **Second term**: $$\int_{0}^{\frac{\pi}{2}} 2\cos(x) \, dx = 2 \int_{0}^{\frac{\pi}{2}} \cos(x) \, dx.$$ The integral of $$\cos(x)$$ is $$\sin(x)$$, so: $$2 \int_{0}^{\frac{\pi}{2}} \cos(x) \, dx = 2 \sin(x) \bigg|_{0}^{\frac{\pi}{2}} = 2\left(\sin\left(\frac{\pi}{2}\right) - \sin(0)\right).$$ Since $$\sin(\frac{\pi}{2}) = 1$$ and $$\sin(0) = 0$$, this becomes: $$2(1 - 0) = 2.$$ --- ### Combine the results Now subtract the two terms: $$\int_{0}^{\frac{\pi}{2}} (2 - 2\cos(x)) \, dx = \pi - 2.$$ --- ### Final Answer: $$\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} 2x \sin(xy) \, dy \, dx = \pi - 2.$$ --- Do you want a breakdown of any specific step or additional examples? Here are 5 related questions to explore further: 1. How does changing the bounds affect the result of a double integral? 2. What is the geometric interpretation of this integral? 3. How can this integral be solved using polar coordinates? 4. What is the role of symmetry in simplifying double integrals? 5. How does integrating with respect to $$x$$ first change the process? **Tip**: Always verify the order of integration to see if switching simplifies the calculations!