Cubic and Trig functions 2

$f(0) = g(0) = 1 \implies d = 1$ and $f(\dfrac{\pi}{2}) = g(\dfrac{\pi}{2}) = -1 \implies \boxed{2\pi^2 b + 4\pi c = -16 -\pi^3}$ and $\int_{0}^{\dfrac{\pi}{2}} g(x) dx$ = $\int_{0}^{\dfrac{\pi}{2}} \cos(x) - \sin(x) \:\ dx = \sin(x) + \cos(x)|_{0}^{\dfrac{\pi}{2}} = 0 \implies$

$0 = \int_{0}^{\dfrac{\pi}{2}} (x^3 + bx^2 + cx + 1) dx = \dfrac{1}{4}x^4 + \dfrac{b}{3}x^3 + \dfrac{c}{2}x^2 + x|_{0}^{\dfrac{\pi}{2}} = \dfrac{\pi^4}{2^6} + \dfrac{\pi^3}{3 * 2^3}b + \dfrac{\pi^2}{2^3}c + \dfrac{\pi}{2} \implies$

$\boxed{8\pi^3b + 24\pi^2c = - 96\pi - 3\pi^4}$

$\boxed{2\pi^2 b + 4\pi c = -16 - \pi^3}$

Solving the above system we obtain:

$c = \dfrac{\pi^3 - 32}{8\pi}$ and $b = -\dfrac{3\pi}{4} \implies$

$\int_{\dfrac{-\pi}{4}}^{0} f(x) dx = \dfrac{x^4}{4} - \dfrac{\pi}{4} x^3 + \dfrac{\pi^3 - 32}{16\pi} x^2 +x|_{\dfrac{-\pi}{4}}^{0} = \dfrac{384\pi - 9\pi^4}{1024}$

and

$\int_{-\dfrac{\pi}{4}}^{0} g(x) dx$ = $\int_{-\dfrac{\pi}{4}}^{0} \cos(x) - \sin(x) \:\ dx = \sin(x) + \cos(x)|_{-\dfrac{\pi}{4}}^{0} = 1$

$\implies \int_{\dfrac{-\pi}{4}}^{0} g(x) - f(x) \:\ dx = \dfrac{9\pi^4 - 384\pi + 1024}{1024} =$ $\dfrac{3^2(\pi^{2})^2 - 3 * 2^7\pi + (2^{2})^{5}}{(2^{2})^5} = \dfrac{\alpha^{\beta}(\pi^{\beta})^{\beta} - \alpha\beta^{\lambda}\pi + (\beta^{\beta})^{\gamma}}{(\beta^{\beta})^{\gamma}} \implies$ $\alpha + \beta + \lambda + \gamma = \boxed{17}$ .