I have tried solving the following integral using Calc I & II methods and couldn't find an answer. I would like to know the techniques for solving such an integral.
$$\int\sqrt{25\sin^2(5x)+49\cos^2(7x)}dx$$
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$\begingroup$ If it does Wolfram doesn't know about it. $\endgroup$– amcaldeCommented Nov 2, 2017 at 15:12
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5$\begingroup$ It is the arc length of a Lissajous curve and it is very likely related to (hyper-)elliptic integrals. What is the original purpose of computing such primitive? $\endgroup$– Jack D'AurizioCommented Nov 2, 2017 at 21:20
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1$\begingroup$ Most likely there is no closed formula, as pointed out in problem 10 in here:math.harvard.edu/~knill/teaching/summer2011/handouts/… $\endgroup$– AndreasCommented Nov 3, 2017 at 9:49
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$\begingroup$ wolframalpha.com/… $\endgroup$– whatamidoingCommented Sep 9 at 16:03
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1 Answer
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$\int\sqrt{25\sin^2(5x)+49\cos^2(7x)}\,dx$ resembles the integral for the arc length of: $$\overrightarrow{r}(t)=\left\langle\cos(5t),\sin(7t) \right\rangle$$
Here $\overrightarrow{r}(t)$ is a Lissajous curve, which is a curve defined as the parametric equations: $$x=A\sin(at+\delta),\;\;y=B\sin(bt)$$ where $t$ is the parameter and $A, a, B, b,\delta$ are constants. We can derive the length of the curve fairly easily: $$L=\int\sqrt{(dx)^2+(dy)^2}=\int\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt=$$ $$\sqrt{\int(Aa\cos(at+\delta))^2+(Bb\cos(bt))^2}\,dt$$
In most cases, including this one, the resulting integral must be solved numerically as there is no closed form solution.