I need to find a function whose arguments are the period of a $\sin$ function and its arc length, and whose output is the height a of the $\sin$ function. In my physical system I need to make a thing which is of given arc length, and what I need to know is the coefficient $a$ in $a\cos(x)$, for a complete cycle of $\cos$ from $0...2\pi$, to satisfy the arc length requirement.
Edit: My physical system contains lines described by $a\sin(bx)$. I need to set $a$ to keep arc length = $2\pi$, for $b<1$ or $b=1$ Attempting to use the Arc Length Formula, I arrived at $$\int_{0}^{2\pi}\sqrt{1+a^2 b^2 \cos^2 (bx)}\,dx$$ My phone is not the ideal instrument on which to learn MathJax well enough to edit my question. Thank you for your patience. Surely there is a better way than fumbling with LaTex on an iphone.
Thank you very much Mark
Your integral is equivalent to $$ab\int_0^{2\pi b}\sqrt{(\sin(bx)+1)}\mathrm dx$$Make a substitution $y=bx$. $$a\int_0^{2\pi b^2}\sqrt{\sin(y)+1}\mathrm dy $$Use $\sin(y)=\cos(\pi/2 - y)$ and $\cos\theta=2\cos^2(\theta/2)-1$. This gives $$a\int_0^{2\pi b^2}\sqrt2\cos\left(\frac\pi4-\frac y2\right)\mathrm dy=-a\sqrt2\left(\sin\left(\frac\pi4-\pi b^2\right)-\sin\frac\pi4\right)=a\sqrt2\left(\frac1{\sqrt2}-\frac1{\sqrt2}\cos(\pi b^2)+\frac1{\sqrt2}\sin(\pi b^2)\right)=a(1-\cos(\pi b^2)+\sin(\pi b^2))$$