I am trying to find the arc length of $y=\sin(ax)+\cos(bx)$ over $\{0\leq x\leq 2\pi\}$ in terms of a and b. With the formula for the arc length I am unable to evaluate the following integral:
$$\int_{0}^{2\pi} \sqrt{1+\bigg( a\cos(ax) -b\sin\big(bx) \bigg) ^{2} } dx $$
I have attempted to use numerous orthodox methods and they have all failed. I suspect it to be some non elementary result in terms of a and b. Any help solving this integral will be much appreciated.
Find the arc length of $y=\sin(ax)+\cos(bx)$
arc lengthcalculusintegration
Best Answer
I could be totally wrong but I think that is $b\neq a$, there no solution to the antiderivative and the integral.
If $b=a$, we have $$I=\int\sqrt{1+\Big( a\cos(ax) -a\sin(ax) \Big) ^{2} } \,dx=\int \sqrt{1+a^2-a^2 \sin (2 a x)}\,dx$$ $$I=-\frac 1a E\left( \frac{\pi }{4}- a x|-2 a^2\right)$$ $$J=\int_0^{2\pi}\sqrt{1+\Big( a\cos(ax) -a\sin(ax) \Big) ^{2} } \,dx$$ $$J=\frac 1a \left(E\left(\frac{\pi }{4}|-2 a^2\right)-E\left(\frac{\pi }{4}-2 a \pi |-2 a^2\right) \right)$$