Quartic functions and Trigonometric functions

$f(0) = g(0) = 0 \implies c = 0$ and $f(\dfrac{\pi}{2}) = g(\dfrac{\pi}{2}) = 1 \implies \boxed{a\pi^4 + 4\pi^2 = 16}$

$\int_{0}^{\dfrac{\pi}{2}} f(x) dx = \int_{0}^{\dfrac{\pi}{2}} \sin(x) dx = 1 \implies$ $1 = \int_{0}^{\dfrac{\pi}{2}} ax^4 + bx^2 dx = \dfrac{ax^5}{5} + \dfrac{bx^3}{3}|_{0}^{\dfrac{\pi}{2}} = \dfrac{\pi^5}{5 * 32}a + \dfrac{\pi^3}{3 * 8}b \implies$

$3\pi^5 a + 20\pi^3 b = 480$

$a\pi^4 + 4\pi^2 = 16$

Solving the above system we obtain: $b = \dfrac{6(10 - \pi)}{\pi^3}$ and $a = \dfrac{40(\pi - 6)}{\pi^5}$

$\implies \int_{0}^{\dfrac{\pi}{4}} f(x) dx = \int_{0}^{\dfrac{\pi}{4}} \dfrac{40(\pi - 6)}{\pi^5}x^4 + \dfrac{6(10 - \pi)}{\pi^3}x^2 dx =$ $\dfrac{8(\pi - 6)}{\pi^5} x^5 + \dfrac{2(10 - \pi}{\pi^3} x^3|_{0}^{\dfrac{\pi}{4}} = \dfrac{\pi - 6}{2^7} + \dfrac{10 - \pi}{2^5} = \dfrac{\pi - 6 + 40 - 4\pi}{128} = \dfrac{34 - 3\pi}{128}$

and,

$\implies \int_{0}^{\dfrac{\pi}{4}} g(x) dx = \int_{0}^{\dfrac{\pi}{4}} \sin(x) dx = -\cos(x)|_{0}^{\dfrac{\pi}{4}} = \dfrac{\sqrt{2} - 1}{\sqrt{2}}$

$\implies \int_{0}^{\dfrac{\pi}{4}} g(x) - f(x) dx = \dfrac{\sqrt{2} - 1}{\sqrt{2}} - (\dfrac{34 - 3\pi}{128}) =$ $\dfrac{\sqrt{2} - 1}{\sqrt{2}} - (\dfrac{2*17 - 3\pi}{2^7}) = \dfrac{\sqrt{\alpha} - \beta}{\sqrt{\alpha}} - (\dfrac{\alpha\lambda - \gamma\pi}{\alpha^{\omega}}) \implies \alpha + \beta + \lambda + \gamma + \omega = \boxed{30}$ .