A calculus problem by Akeel Howell

$\begin{aligned} I & = \int_0^\pi x^4 \sin x \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = -x^4 \cos x \bigg|_0^\pi + \int_0^\pi 4 x^3 \cos x \ dx & \small \color{#3D99F6} \text{By IBP again} \\ & = \pi^4 + 4x^3 \sin x \bigg|_0^\pi - \int_0^\pi 12 x^2 \sin x \ dx & \small \color{#3D99F6} \text{By IBP again} \\ & = \pi^4 + 0 + 12x^2 \cos x \bigg|_0^\pi - \int_0^\pi 24 x \cos x \ dx & \small \color{#3D99F6} \text{By IBP again} \\ & = \pi^4 - 12\pi^2 - 24 x \sin x \bigg|_0^\pi + \int_0^\pi 24 \sin x \ dx & \small \color{#3D99F6} \text{By IBP again} \\ & = \pi^4 - 12\pi^2 - 24 \cos x \bigg|_0^\pi \\ & = \pi^4 - 12\pi^2 + 48 \end{aligned}$

$\implies a+b+c+d = 4+12+2+48 = \boxed{66}$

Using a repeated sequence of integration-by-parts, we obtain:

$-x^{4}cos(x) + 4x^{3}sin(x) - 12x^{2}cos(x) + 24xsin(x) - 24cos(x)$

which evaluating the bounds $x = 0, \pi$ gives:

$\pi^{4} - 12\pi^{2} + 48 \Rightarrow a = 4, b = 12, c = 2, d = 48 \Rightarrow a + b + c + d = \boxed{66}.$