Question

Find the ∫∫x^3 y^3 sin⁡(x)sin⁡(y) dxdy.

(a) (x^3 Cos(x) + 3x^2 Sin(x) + 6xCos(x)-6Sin(x))(y^3 Cos(y) + 3[y^2 Sin(y) – 2[-yCos(y) + Sin(y)]])

(b) (-x^3 Cos(x) – 3x^2 Sin(x) – 6xCos(x)-6Sin(x))(-y^3 Cos(y) + 3[y^2 Sin(y) – 2[-yCos(y) + Sin(y)]])

(c) (-x^3 Cos(x) + 3x^2 Sin(x) + 6xCos(x)-6Sin(x))(-y^3 Cos(y) + 3y^2 Sin(y) + 6yCos(y) – 6Sin(y))

(d) (–x^3 Cos(x) + 6xCos(x) – 6Sin(x))(-y^3 Cos(y))

I got this question in a job interview.

This interesting question is from Double Integrals topic in section Multiple Integrals of Engineering Mathematics

Answer 1

The correct option is (c) (-x^3 Cos(x) + 3x^2 Sin(x) + 6xCos(x)-6Sin(x))(-y^3 Cos(y) + 3y^2 Sin(y) + 6yCos(y) – 6Sin(y))

For explanation I would say: Add constant automatically

∫x^3 Sin(x)dx = -x^3 Cos(x) + 3∫x^2 Cos(x)dx

∫x^2 Cos(x)dx = x^2 Sin(x) – 2∫xSin(x)dx

∫xSin(x)dx = -xCos(x) + ∫Cos(x)dx = -xCos(x) + Sin(x)

=> ∫x^3 Sin(x)dx = -x^3 Cos(x) + 3[x^2 Sin(x) – 2[-xCos(x) + Sin(x)]]

=> ∫x^3 Sin(x)dx = -x^3 Cos(x) + 3x^2 Sin(x) + 6xCos(x) – 6Sin(x)

and, ∫y^3 Sin(y)dy = -y^3 Cos(x) + 3∫y^2 Cos(y)dy

∫y^2 Cos(y)dy = y^2 Sin(y) – 2∫ySin(y)dy

∫ySin(y)dy = -yCos(y) + ∫Cos(y)dy = -yCos(y) + Sin(y)

=> ∫y^3 Sin(y)dy = -y^3 Cos(y) + 3[y^2 Sin(y) – 2[-yCos(y) + Sin(y)]]

=> ∫y^3 Sin(y)dy = -y^3 Cos(y) + 3y^2 Sin(y) + 6yCos(y) – 6Sin(y)

Hence, ∫∫x^3 y^3 sin⁡(x) sin⁡(y) dxdy = (∫x^3 Sin(x)dx)(∫y^3 Sin(y)dy) = (-x^3 Cos(x) + 3x^2 Sin(x)+6xCos(x) – 6Sin(x))(-y^3 Cos(y) + 3y^2 Sin(y) + 6yCos(y) – 6Sin(y)).