Find the ∫∫x3 y3 sin(x)sin(y) dxdy. ]) b) (-x3 Cos(x) – 3x2 Sin(x) – 6xCos(x)-6Sin(x))(-y3 Cos(y) + 3]) c) (-x3 Cos(x) + 3x2 Sin(x) + 6xCos(x)-6Sin(x))(-y3 Cos(y) + 3y2 Sin(y) + 6yCos(y) – 6Sin(y)) d) (–x3 Cos(x) + 6xCos(x) – 6Sin(x))(-y3 Cos(y))
Correct Answer: y2 Sin(y) – 2[-yCos(y) + Sin(y)
Add constant automatically ∫x3 Sin(x)dx = -x3 Cos(x) + 3∫x2 Cos(x)dx ∫x2 Cos(x)dx = x2 Sin(x) – 2∫xSin(x)dx ∫xSin(x)dx = -xCos(x) + ∫Cos(x)dx = -xCos(x) + Sin(x) => ∫x3 Sin(x)dx = -x3 Cos(x) + 3] => ∫x3 Sin(x)dx = -x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x) and, ∫y3 Sin(y)dy = -y3 Cos(x) + 3∫y2 Cos(y)dy ∫y2 Cos(y)dy = y2 Sin(y) – 2∫ySin(y)dy ∫ySin(y)dy = -yCos(y) + ∫Cos(y)dy = -yCos(y) + Sin(y) => ∫y3 Sin(y)dy = -y3 Cos(y) + 3] => ∫y3 Sin(y)dy = -y3 Cos(y) + 3y2 Sin(y) + 6yCos(y) – 6Sin(y) Hence, ∫∫x3 y3 sin(x) sin(y) dxdy = (∫x3 Sin(x)dx)(∫y3 Sin(y)dy) = (-x3 Cos(x) + 3x2 Sin(x)+6xCos(x) – 6Sin(x))(-y3 Cos(y) + 3y2 Sin(y) + 6yCos(y) – 6Sin(y)).