# Periodic functions for the definite integral

Considering this question where there is this integral:

$$int_{0}^{4pi} ln|13sin x+3sqrt3 cos x|mathrm dx tag 1$$

Easily all the periodic function $$a’sin(x)+bcos(x)+c=0 tag 2$$ can be written as:

$$Asin(x+phi)+c=0, A=sqrt{a’^2+b^2}quad text{ or }quad Acos(x+varphi)+c=0tag 3$$
where $$phi, varphi=arctan ldots$$ are angles definited in radians, hence $$inBbb R$$.
Reading the comments of the user @Sangchul Lee, I think that $$|sin(x)|$$ is a even-function and $$pi-$$periodic,

$$int_{0}^{4pi} ln|13sin x+3sqrt3 cos x|,mathrm{d}x=4int_{0}^{pi}ln| Asin(x+phi)|,mathrm{d}x=4int_{0}^{pi}ln(A| sin(x+phi)|),mathrm{d}x$$

1. Why $$phi$$ vanished? It is true if $$phi=Kpi$$, with $$KinBbb Z$$. I not remember this now.
2. Considering the comment "Let $$f:mathbb R→mathbb R$$ be $$T$$-periodic and integrable on any finite interval then
$$∫_0^Tf(x)dx=∫_0^Tf(x+a)dx$$" when is it useful, for a periodic function,

$$int_{0}^{T}f(x),mathrm{d}x=int_{0}^{T}f(x+a),mathrm{d}x=int_{color{red}{-a}}^{color{red}{T-a}}f(x+a),mathrm{d}x$$

and if are there general rules (or what is it happen) for the limits of the integral of a generic periodic function?

$$int_{color{blue}{lambda}}^{color{blue}{mu}}f(x+a),mathrm{d}x=int_{color{blue}{lambda}}^{color{blue}{mu}}f(x),mathrm{d}x=Cint_{color{magenta}{cdots}}^{color{magenta}{cdots}}f(x),mathrm{d}x$$
where $$C=C(lambda)$$ (upper bound) or $$C=C(mu)$$ (lower bound) is a real constant.

Mathematics Asked on November 21, 2021

We have $$13^2+27=14^2$$, so $$13sin(x)+3sqrt{3}cos(x) = 14sin(x+varphi),qquad varphi=arctanfrac{3sqrt{3}}{13}notinpimathbb{Z}$$ and $$begin{eqnarray*}int_{0}^{4pi}logleft|13sin(x)+3sqrt{3}cos(x)right|,dx &=& 4pilog(14)+int_{0}^{4pi}logleft|sin xright|,dx\&=&4pilog(14)+4int_{0}^{pi}logsin(x),dxend{eqnarray*}$$ where $$begin{eqnarray*} I=int_{0}^{pi}logsin(x),dx &=& 2int_{0}^{pi/2}logsin(2z),dz\&=&pilog(2)+2int_{0}^{pi/2}logsin(z),dz+2int_{0}^{pi/2}logcos(z),dz\&=&pilog(2)+2Iend{eqnarray*}$$ leads to $$int_{0}^{4pi}logleft|13sin(x)+3sqrt{3}cos(x)right|,dx =4pilog(14)-4pilog(2) = color{red}{4pilog(7)}.$$

Answered by Jack D'Aurizio on November 21, 2021

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