Something my AP Calculus class does not teach: Real-Complex Integral?!

Relevant wiki: Euler's Formula

$\begin{aligned} f(x) & = \sum_{p=0}^n a_p ({\color{#3D99F6}\cos x + i \sin x})^p & \small \color{#3D99F6} \text{By Euler's formula }e^{i\theta} = \cos \theta + i\sin \theta \\ & = \sum_{p=0}^n a_p ({\color{#3D99F6}e^{ix}})^p \\ & = \sum_{p=0}^n a_p e^{ipx} \\ |f(x)|^2 & = \sum_{p=0}^n \left(a_p e^{ipx}\right)\left(a_p e^{-ipx}\right) & \small \color{#3D99F6} \text{Since }|z|^2 = z\bar z \text{, where }\bar z\text{ is the conjugate of }z. \\ & = \sum_{p=0}^n a_p^2 \end{aligned}$

$\begin{aligned} \implies \int_0^{2\pi} |f(x)|^2 dx & = \int_0^{2\pi} \sum_{p=0}^n a_p^2 \ dx = 2\pi \sum_{p=0}^n a_p^2 \end{aligned}$

$\implies C = 2\pi \approx \boxed{6.183}$

The catch is to use Euler's formula to nicely rewrite $\cos x+ i \sin x$ as $e^{ix}$ . Then, $f$ becomes $f(x) = \sum_{p=0}^n a_p e^{ipx} .$ Since $a_p \in \mathbb{R}$ , $|f(x)|^2 = f(x) \overline{f(x)} = \left(\sum_{p=0}^n a_p e^{ipx} \right) \left( \sum_{q=0}^n a_q e^{-iqx} \right) = \sum_{p,q = 0}^n a_p a_q e^{i(p-q)x}$

To integrate $|f(x)|^2$ , we use the following identities that can easily be proven by standard calculus: $\int_0^{2\pi} \cos(rx) dx = \int_0^{2\pi} \sin(sx) dx = 0$ , for all non-zero integers $r$ and $s$ . Because the complex exponential is just a linear combination of a sine and a cosine, we have, for $r \in \mathbb{Z}$ , $\int_0^{2\pi} e^{irx} dx = \begin{cases} 0, & \text{if } r \neq 0 \\ 2\pi, & \text{if } r = 0 \end{cases}$

Finally, $\int_0^{2\pi} |f(x)|^2 dx = \sum_{p,q = 0}^n a_p a_q \int_0^{2\pi} e^{i(p-q)x} dx = 2\pi \sum_{p=0}^n a_p^2,$ where, in the last equation, we only summed the terms with $p=q$ , whose integrals are non-zero. Therefore, $C = 2\pi \approx 6.28$

Assume that the problem statement is valid for any $n$ and any $a_p$ (it should be, otherwise it would be unsolvable). Then we can just set $n=1$ , $a_0 = 1, a_1 = 0$ , and get

$\int_0^{2 \pi} |f(x)|^2 dx = \int_0^{2 \pi} 1 dx = 2 \pi = C$