Quartic functions and Trigonometric functions 2

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

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

$\boxed{3\pi^5a + 20\pi^3b = 960 - 240\pi}$

$\boxed{\pi^4a + 4\pi^2b = 0}$

Solving the above system we obtain:

$a = 120(\dfrac{\pi - 4}{\pi^5})$ and $b = 30(\dfrac{4 - \pi}{\pi^3}) \implies$

$\int_{\dfrac{-\pi}{4}}^{0} f(x) dx = \int_{\dfrac{-\pi}{4}}^{0} (120(\dfrac{\pi - 4}{\pi^5})x^4 + 30(\dfrac{4 - \pi}{\pi^3})x^2 + 1) dx = 24(\dfrac{\pi - 4}{\pi^5})x^5 + 10(\dfrac{4 - \pi}{\pi^3})x^3 + x|_{\dfrac{-\pi}{4}}^{0} =$

$\dfrac{6}{4^4}(\pi - 4) + \dfrac{10}{4^3}(4 - \pi) + \dfrac{\pi}{4} = \dfrac{6(\pi - 4) + 40(4 - \pi) + 64\pi}{256} = \dfrac{30\pi + 136}{256} = \dfrac{15\pi + 68}{128}$

and,

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

$\implies \int_{\dfrac{-\pi}{4}}^{0} f(x) - g(x) dx = \dfrac{15\pi + 68}{128} - (\sqrt{2} - 1) = \dfrac{196}{128} + \dfrac{15\pi}{128} - \sqrt{2} = \dfrac{49}{32} + \dfrac{15\pi}{128} - \sqrt{2} =$ $\dfrac{7^2}{2^5} + \dfrac{3 * 5\pi}{2^7} - \sqrt{2} = \dfrac{\alpha^{\beta}}{\beta^{\lambda}} + \dfrac{\gamma * \lambda\pi}{ \beta^{\alpha}} - \sqrt{\beta} \implies$ $\alpha + \beta + \lambda + \gamma = \boxed{17}$ .