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$$

- Why $phi$ vanished? It is true if $phi=Kpi$, with $KinBbb Z$. I not remember this now.
- 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

1 AnswersWe 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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