I I y^2*cos2xy dxdy D: x=0, y=sqrt п/2, y=x/2
Помогите пожалуйста...
int int (D) y^2 * cos (2xy) dx dy, D: x = 0, y = pi^(1/2)/2, y = x/2.
Проинтегрируем интеграл сначала по у, потом по х.
y = x/2 => x = 2 * y.
Значит 0 <= x <= 2 * y.
x = 0, y = x/2 => y = 0 => 0 <= y <= pi^(1/2)/2
Тогда
int int (D) y^2 * cos (2xy) dx dy =
= int (0 pi^(1/2)/2) dy int (0 2 * y) y^2 * cos (2xy) dx =
= int (0 pi^(1/2)/2) dy (y^2 * 1/(2 * y) * sin (2xy))_{0}^{2 * y} =
= int (0 pi^(1/2)/2) y/2 * (sin (4 * y^2) - sin 0) dy =
= int (0 pi^(1/2)/2) y/2 * sin (4 * y^2) dy =
= 1/2 * int (0 pi^(1/2)/2) y * sin (4 * y^2) dy =
= 1/2 * int (0 pi^(1/2)/2) sin (4 * y^2) d(1/2 * y^2) =
= 1/2 * 1/2 * int (0 pi^(1/2)/2) sin (4 * y^2) d(y^2) = | y^2 = t | =
= 1/4 * int (0 pi/4) sin (4 * t) dt = 1/4 * (-1/4 * cos (4 * t))_{0}^{pi/4} =
= 1/4 * (-1/4 * cos pi + 1/4 * cos 0) = 1/4 * (1/4 + 1/4) = 1/4 * 1/2 = 1/8.
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