Quartic Imitation Cosine

$g(0)=\cos0=1$ , so the quartic equation passes through (0,1). Therefore it is in the form $ax^4+bx^2+1$ . Now we need two equations in order to solve simultaneously.

First, we use the fact that it cuts the x-axis at $\frac{\pi}{2}\,(g(\frac{\pi}{2})=\cos\frac{\pi}{2}=\frac{\pi}{2})$ : $\text{When }x=\frac{\pi}{2},y=0:\\\frac{a\pi^4}{16}+\frac{b\pi^2}{4}+1=0$ Next, we use the fact that the areas under f(x) and g(x) are equal between $\frac{\pi}{2}$ and $-\frac{\pi}{2}$ : $\begin{aligned}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(ax^4+bx^2+1\right)dx&=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}(\cos x)dx\\ 2\int_0^\frac{\pi}{2}\left(ax^4+bx^2+1\right)dx&=2\int_0^\frac{\pi}{2}(\cos x)dx&\text{(as both functions are even)}\\ \left[\frac{ax^5}{5}+\frac{bx^3}{3}+x\right]_0^\frac{\pi}{2}&=\left[\sin x\right]_0^\frac{\pi}{2}\\ \frac{a\pi^5}{160}+\frac{b\pi^3}{24}+\frac{\pi}{2}&=1\end{aligned}$ Solving these equations simultaneously, we get: $a=\frac{80}{\pi^4}-\frac{240}{\pi^5}\qquad\qquad b=\frac{60}{\pi^3}-\frac{24}{\pi^2}\\ f(x)=\left(\frac{80}{\pi^4}-\frac{240}{\pi^5}\right)x^4+\left(\frac{60}{\pi^3}-\frac{24}{\pi^2}\right)x^2+1$ Now, the difference between the area under f(x) and the area under g(x) between between $\frac{\pi}{4}$ and $-\frac{\pi}{4}$ is: $\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\left(\left(\frac{80}{\pi^4}-\frac{240}{\pi^5}\right)x^4+\left(\frac{60}{\pi^3}-\frac{24}{\pi^2}\right)x^2+1-\cos x\right)dx\\ =0.0006093714$