Verified answer

∫x^3dx - ∫xcos^2(x)dx

x^4/4 - ∫xcos^2(x)dx

cos^2(x) = (1+cos(2x))/2

x^4/4 - 1/2∫x(1+cos(2x))dx

x^4/4 - 1/2 ∫(x+xcos(2x))dx

x^4/4 - 1/4x^2 - 1/2 ∫(xcos(2x))dx

Let u=x, and dv = cos(2x)

du = dx and v = 1/2 sin(2x)

∫udv = uv - ∫vdu

∫xcos(2x)dx = x/2sin(2x) - 1/2∫sin(2x)dx

∫xcos(2x)dx = x/2sin(2x) + 1/4cos(2x)

x^4/4 - 1/4x^2 - 1/2 [x/2sin(2x) + 1/4cos(2x)]

(1/4)x^4 - (1/4)x^2 - x/4sin(2x) - 1/8cos(2x)+C

I = ʃx^3dx - ʃx*cos^2(x)dx

First integral is straightforward: I_1 = ʃx^3dx = x^4/4

second integral we do by parts:

u = x, du = 1 ,

dv = cos^2(x) , v = ʃcos^2(x) = ʃ0.5cos(2x) + 0.5 dx = 0.25sin(2x) + 0.5x

the formula we need is:

I_2 = uv - ʃv*du

I_2 = x[0.25sin(2x) + 0.5] - ʃ0.25sin(2x) + 0.5x dx

I_2 = 1/8 * [ 2x^4 + 2xsin(2x) + cos(2x) + 2x^2]

hence the integral of the whole function is:

I = I_1 - I_2 = x^4/4 - 1/8 * [ 2x^4 + 2xsin(2x) + cos(2x) + 2x^2] + c