Riley (3rd ed., Mathematical Methods of Physics,p420) goes on stating that for an even or odd symmetry of $f(x)$ about the quarter period $f(\frac{L}{4}-x) = \pm f(x-\frac{L}{4})$ we can write $$b_r = \frac{2}{L} \int^{x_0+L}_{x_0}f(s)\sin\bigg(\frac{2\pi rs}{L}+\color{red}{\frac{\pi r}{2}}\bigg)\mathrm{ds},$$ after the substitution $s=x-\frac{L}{4}$.
I can't make any sense of the $\frac{\pi r}{2}$ term inside the sine argument. This is problematic because we go on expanding the sine in order to arrive at a certain result. It seems far fetched that this is a typo because the result depends on it.
Where could the $\frac{\pi r}{2}$ term come from?
I'll derive a similar result, as I've exhausted the ways in which to explain the stated relation in the question; I'll assume that it must be a typo in the text.
Consider a function $f$ which has a parity about its quarter period. We want to exploit this parity to make the Fourier coefficient look simpler.
We look into the case of functions $f$ that have odd parity with respect to the origin.
Since $f$ has a quarter period symmetry, we want to apply a change of variables $s = x-L/4$; in order to see what kind of effect this has on the trigonometric part of the integrand.
$$b_r = \frac{2}{L}\int_{x_0}^{x_0 +L}f(s)\sin\bigg(\frac{2\pi rs}{L}\bigg)ds \ . $$ Note that, since $dx=ds$, the result is invariant with respect to integration variable.
$$b_r = \frac{2}{L}\int_{x_0}^{x_0 +L}f(s)\sin\bigg(\frac{2\pi r(x-L/4)}{L}\bigg)dx \\ =\frac{2}{L}\int_{x_0}^{x_0 +L}f(s)\sin\bigg(\frac{2\pi rx}{L} - \color{red}{\frac{\pi r}{2}}\bigg)dx \ .$$
Expanding the sine,
$$ \sin\bigg(\frac{2\pi rx}{L} - \color{red}{\frac{\pi r}{2}}\bigg) = \color{green}{\cos\bigg(\frac{\pi r}{2}\bigg)}\sin\bigg(\frac{2\pi rx}{L}\bigg)+ \color{green}{\sin\bigg(\frac{\pi r}{2}\bigg)}\cos\bigg(\frac{2\pi rx}{L}\bigg) $$ , we see that either $b_{2r}$ or $b_{2r+1}$ vanishes depending on the parity of $f$ about $L/4$.