I am currently studying A Student's Guide to the Schrödinger Equation by Daniel A. Fleisch. Chapter 1.6 Finding Components Using the Inner Product says the following:
To see how this works, consider a function $\mid \psi(x) \rangle$ expanded using the basis functions $\mid \psi_1 \rangle = \sin(x)$, $\mid \psi_2 \rangle = \cos(x)$, $\mid \psi_3(x) \rangle = \sin(2x)$ over the interval $x = -\pi$ to $x = \pi$: $$\psi (x) = 5 \mid \psi_1 (x) \rangle - 10 \mid \psi_2(x) \rangle + 4 \mid \psi_3(x) \rangle.$$ In this case, you can read the components $c_1 = 5$, $c_2 = -10$, $c_3 = 4$ directly from this equation for $\psi(x)$. But to understand how Eq. 1.36 gives these values, write $$c_1 = \dfrac{\int^{\infty}_{-\infty} \psi_1^*(x) \psi(x) \ dx}{\int_{-\infty}^{\infty} \psi_1^* (x) \psi_1(x) \ dx} = \dfrac{\int_{-\pi}^{\pi} [\sin(x)]^*[5\sin(x) - 10\cos(x) + 4\sin(2x)] \ dx}{\int_{-\pi}^{\pi} [\sin(x)]^* [\sin(x)] \ dx}$$ $$c_2 = \dfrac{\int^{\infty}_{-\infty} \psi_2^*(x) \psi(x) \ dx}{\int_{-\infty}^{\infty} \psi_2^* (x) \psi_2(x) \ dx} = \dfrac{\int_{-\pi}^{\pi} [\cos(x)]^*[5\sin(x) - 10\cos(x) + 4\sin(2x)] \ dx}{\int_{-\pi}^{\pi} [\cos(x)]^* [\cos(x)] \ dx}$$ $$c_3 = \dfrac{\int^{\infty}_{-\infty} \psi_3^*(x) \psi(x) \ dx}{\int_{-\infty}^{\infty} \psi_3^* (x) \psi_3(x) \ dx} = \dfrac{\int_{-\pi}^{\pi} [\sin(2x)]^*[5\sin(x) - 10\cos(x) + 4\sin(2x)] \ dx}{\int_{-\pi}^{\pi} [\sin(2x)]^* [\sin(2x)] \ dx}$$ These integrals can be evaluated with the help of the relations
$$\int_{-\pi}^\pi \sin^2(ax) \ dx = \left[ \dfrac{x}{2} - \dfrac{\sin(2ax)}{4a} \right]^\pi_{-\pi} = \pi$$ $$\int_{-\pi}^\pi \cos^2(ax) \ dx = \left[ \dfrac{x}{2} - \dfrac{\sin(2ax)}{4a} \right]^\pi_{-\pi} = \pi$$ $$\int_{-\pi}^\pi \sin(x)\cos(x) \ dx = \left[ \dfrac{1}{2} \sin^2(x) \right]^\pi_{-\pi} = 0$$ $$\int_{-\pi}^\pi \sin(mx) \sin(nx) \ dx = \left[ \dfrac{\sin(m - n)x}{2(m - n)} + \dfrac{\sin(m + n)x}{2(m + n)} \right]^\pi_{-\pi} = 0$$ in which $m$ and $n$ are (different) integers. Applying these gives $$c_1 = \dfrac{5(\pi) - 10(0) + 4(0)}{\pi} = 5$$ $$c_2 = \dfrac{5(0) - 10(\pi) + 4(0)}{\pi} = -10$$ $$c_3 = \dfrac{5(0) - 10(0) + 4(\pi)}{\pi} = 4$$ as expected. Notice that in this example the basis functions $\sin(x)$, $\cos(x)$, and $\sin(2x)$ are orthogonal but not orthonormal, since their norms are $\pi$ rather than one. Some students express surprise that sinusoidal functions are not normalized, since their values run from $−1$ to $+1$. But remember that it’s the integral of the square of the function, not the peak value, that determines the function’s norm.
It's this part that I am confused about:
...as expected. Notice that in this example the basis functions $\sin(x)$, $\cos(x)$, and $\sin(2x)$ are orthogonal but not orthonormal, since their norms are $\pi$ rather than one. Some students express surprise that sinusoidal functions are not normalized, since their values run from $−1$ to $+1$. But remember that it’s the integral of the square of the function, not the peak value, that determines the function’s norm.
How are the norms of $\sin(x)$, $\cos(x)$, and $\sin(2x)$ equal to $\pi$? Such a statement doesn't even make sense, since these are (complex) functions -- not scalars. For instance, the norm of $\sin(x)$, as an element of a Hilbert space, is $$|\sin(x)| = \langle \sin(x) \mid \sin(x) \rangle = \sqrt{\int_{-\infty}^\infty \sin^*(x) \sin(x) \ dx} = \sqrt{\int_{-\infty}^\infty \sin^2(x) \ dx}.$$ The author even makes the incorrect statement that "...it’s the integral of the square of the function, not the peak value, that determines the function’s norm". I would greatly appreciate it if people would please take the time to clarify this section.
And since the magnitude (or norm) of a vector $\vec{A}$ can be found by taking the square root of the inner product of the vector with itself, the complex conjugate is built into the process of taking the inner product between complex quantities: $$|A| = \sqrt{\vec{A} \cdot \vec{A}} = \sqrt{A_x^* A_x + A_y^* A_y + A_z^* A_z} = \sqrt{\sum_{i = 1}^N} A_i^* A_i. \tag{1.26}$$ This also applies to complex functions: $$|f(x)| = \sqrt{\langle f(x) \mid f(x) \rangle} = \sqrt{\int_{-\infty}^\infty f^*(x) f(x) \ dx}. \tag{1.27}$$ So it’s necessary to use the complex conjugate to find the norm of a complex vector or function.