Question

Вычислить двойной интеграл

Answer 1

$\displaystyle \int{\int\limits_{(D)} {sin(x+y)} \, dx } \, dy=\int\limits^{\frac{\pi}{2} }_ {0} \, dx \int\limits^{\frac{\pi}{2}}_x {sin(x+y)} \, dy$

$\displaystyle \int\limits^{\frac{\pi}{2}}_x {sin(x+y)} \, dy = \int\limits^{\frac{\pi}{2}}_x {sin(x+y)} \, d(y+x) = -cos(x+y) ~~~\bigg|^{\frac{\pi}{2}}_x = -cos(x+\frac{\pi}{2})+cos(2x)$

$-cos(x+\frac{\pi}{2}) = -*(-sin(x)) = sin(x)$

$\displaystyle \int\limits^{\frac{\pi}{2}}_0 {cos(2x)+sin(x)} \, dx = \displaystyle \int\limits^{\frac{\pi}{2}}_0 {cos(2x)}\,dx + \int\limits^{\frac{\pi}{2}}_0 sin(x)}\,dx \ \ \ : \\\\\\ 1)~~~~\frac{1}{2} \int\limits^{\frac{\pi}{2}}_0 {cos(2x)}\,d(2x) = \frac{1}{2}sin(2x)~~\bigg|^{\frac{\pi}{2}}_0 = \frac{1}{2}( sin(\pi)-sin(o)) = \frac{1}{2}*0 = 0 \\\\\\2)~~\int\limits^{\frac{\pi}{2}}_0 sin(x)}\,dx = -cos(x)~~\bigg|^{\frac{\pi}{2}}_0 = -(cos(\frac{\pi}{2})-cos(0)) = -1*(0-1) = 1$

ответ: 1