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Iterated Integrals and Area (Exercises)
https://math.libretexts.org/Courses/Georgia_State_University_-_Perimeter_College/MATH_2215%3A_Calculus_III/15%3A_Multiple_Integration/Iterated_Integrals_and_Area_(Exercises)
(b) $\displaystyle \int_0^{\pi} \int_0^{y} (\cos x \sin y)\,dx\,dy\quad$ $=\quad \displaystyle \int_{0}^{\pi} \sin^2 y \, dy\quad$ \(=\quad \displaystyle \int_{0}^{\pi} \frac{1-\cos 2y}{2}\, dy\qu...(b) $\displaystyle \int_0^{\pi} \int_0^{y} (\cos x \sin y)\,dx\,dy\quad$ $=\quad \displaystyle \int_{0}^{\pi} \sin^2 y \, dy\quad$ $=\quad \displaystyle \int_{0}^{\pi} \frac{1-\cos 2y}{2}\, dy\quad$ $=\quad \dfrac{y - \frac{1}{2}\sin 2y}{2}\bigg|_{0}^{\pi}\quad$ $=\quad \frac{\pi}{2} - 0\quad$ $=\quad \dfrac{\pi}{2}$
Iterated Integrals and Area (Exercises)
https://math.libretexts.org/Courses/Montana_State_University/M273%3A_Multivariable_Calculus/15%3A_Multiple_Integration/Iterated_Integrals_and_Area/Iterated_Integrals_and_Area_(Exercises)
(b) $\displaystyle \int_0^{\pi} \int_0^{y} (\cos x \sin y)\,dx\,dy\quad$ $=\quad \displaystyle \int_{0}^{\pi} \sin^2 y \, dy\quad$ \(=\quad \displaystyle \int_{0}^{\pi} \frac{1-\cos 2y}{2}\, dy\qu...(b) $\displaystyle \int_0^{\pi} \int_0^{y} (\cos x \sin y)\,dx\,dy\quad$ $=\quad \displaystyle \int_{0}^{\pi} \sin^2 y \, dy\quad$ $=\quad \displaystyle \int_{0}^{\pi} \frac{1-\cos 2y}{2}\, dy\quad$ $=\quad \dfrac{y - \frac{1}{2}\sin 2y}{2}\bigg|_{0}^{\pi}\quad$ $=\quad \frac{\pi}{2} - 0\quad$ $=\quad \dfrac{\pi}{2}$
4.2: Iterated Integrals and Area (Exercises)
https://math.libretexts.org/Courses/Misericordia_University/MTH_226%3A_Calculus_III/Chapter_15%3A_Multiple_Integration/4.02%3A_Iterated_Integrals_and_Area_(Exercises)
(b) $\displaystyle \int_0^{\pi} \int_0^{y} (\cos x \sin y)\,dx\,dy\quad$ $=\quad \displaystyle \int_{0}^{\pi} \sin^2 y \, dy\quad$ \(=\quad \displaystyle \int_{0}^{\pi} \frac{1-\cos 2y}{2}\, dy\qu...(b) $\displaystyle \int_0^{\pi} \int_0^{y} (\cos x \sin y)\,dx\,dy\quad$ $=\quad \displaystyle \int_{0}^{\pi} \sin^2 y \, dy\quad$ $=\quad \displaystyle \int_{0}^{\pi} \frac{1-\cos 2y}{2}\, dy\quad$ $=\quad \dfrac{y - \frac{1}{2}\sin 2y}{2}\bigg|_{0}^{\pi}\quad$ $=\quad \frac{\pi}{2} - 0\quad$ $=\quad \dfrac{\pi}{2}$

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