Chapter 14 - Multiple Integrals - 14.1 Double Integrals - Exercises Set 14.1 - Page 1007: 6

$$\int_{0}^{1} \int_{0}^{2} y \sin x \ d x \ d y=\frac{ 1-\cos 2}{2}$$,

Given $$\int_{0}^{1} \int_{0}^{2} y \sin x \ d x \ d y$$ So, we get \begin{aligned}I &=\int_{0}^{1} \int_{0}^{2} y \sin x \ d x \ d y\\ &=\int_{0}^{2} \sin x \ d x \int_{0}^{1} y \ d y \\ &=[-\cos x]_{0}^{2}\left[\frac{y^{2}}{2}\right]_{0}^{1} \\ &=\frac{ -\cos 2+1}{2}\\ \end{aligned}