# Can it possible to calculate this integral directly, probably concerning ellipti integral

A student asked me to help her calculating this problem: Assume the length $L$ of a curve is given, and the equation of the curve is \begin{gather*} y(x)=A \sin\Big(\pi x-\frac{\pi}{2}\Big), 0\leq x\leq 1, \end{gather*} calculate the amplitude $A$, where $A\in\mathbb{R}$, actualy you can assume that $A$ is a positive real number. Here is my incomplete solution:

$\because y=-A\cos(\pi x), y'(x)=\pi A \sin (\pi x)=k\sin(\pi x), \text{ where } k=\pi A,$ therefore, \begin{align*} L&=\int_0^1\sqrt{1+(y'(x))^2} dx=\frac{1}{\pi}\int_0^1\sqrt{1+k^2\sin^2(\pi x)}d(\pi x)=\frac{1}{\pi}\int_0^{\pi}\sqrt{1+k^2\sin^2(s)}ds\\ &=\frac{2}{\pi}\int_0^{\pi/2}\sqrt{1+k^2\sin^2(s)}ds=\frac{2}{\pi}\int_0^1\frac{\sqrt{1+k^2 t^2}}{\sqrt{1-t^2}}dt, \quad (\sin s=t). \end{align*}

Then I do not know how to proceed, because I know the last integral is the complete Ellptic Integral EllipticE($\sqrt{-\pi^2A^2}$), as indicated in Maple 16.

After using Maple, I found that it is very difficult to solve for $A$ through equation \begin{gather*} EllipticE(\sqrt{-\pi^2A^2})=\pi L/2. \end{gather*} But after I searched this website, I found the thread How do I integrate the following?. Then I was very doubt that the result of Maple. Can it be possible to calculate the $A$ from the length $L$ analytically, or is it possible to evaluate the integral $\int_0^1\sqrt{1+\pi^2A^2\sin^2(\pi x)}dx$ directly, without the help of Elliptic Integral?

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I may see how to do the indefinite integral, but it's going to be messy... –  Mike Nov 3 '12 at 6:44
I'm confused. Isn't the curve $(x,y(x)) = (x, A \sin(\pi x - n/2))$? So its length would involve an integral of cos, right? –  Gunnar Magnusson Nov 3 '12 at 7:55

Let me try this from your last step.

$$t=\frac{\tan\theta}k,dt=\frac{\sec^2\theta d\theta}{k}$$ $$\frac2\pi\int\limits_0^1\frac{\sqrt{1+k^2 t^2}}{1-t^2}dt=\frac2\pi\int\limits_0^{\tan^{-1}k}\frac{\frac{\sec^2\theta}k\sqrt{1+\tan^2\theta}d\theta}{1-\frac{\tan^2\theta}{k^2}}=$$ $$\frac{2k}\pi\int\limits_0^{\tan^{-1}k}\frac{\sec^3\theta d\theta}{k^2-\tan^2\theta}=2A\int\limits_0^{\tan^{-1}k}\frac{\cos\theta d\theta}{\cos^2\theta(k^2\cos^2\theta-\sin^2\theta)}=$$ $$2A\int\limits_0^{\tan^{-1}k}\frac{\cos\theta d\theta}{(1-\sin^2\theta)(k^2-k^2\sin^2\theta-\sin^2\theta)}$$ Partial fractions time.

$$u=\sin\theta,du=\cos\theta d\theta$$ $$2A\int\limits_0^{\tan^{-1}k}\frac{\cos\theta d\theta}{(1-\sin^2\theta)(k^2-k^2\sin^2\theta-\sin^2\theta)}=$$ $$2A\int\limits_1^\frac1{\sqrt{k^2+1}}\frac{du}{(1-u)(1+u)(k-u\sqrt{k^2+1})(k+u\sqrt{k^2+1})}$$

If I've done everything right so far, this should lead to an answer. I may work this further later to make sure I get a matching answer.

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Mike, I am very sorry for my mistyping in the integrand of the last integral, it should be $\frac{sqrt{1+k^2 t^2}{\sqrt{1-t^2}}$. –  azhi Nov 3 '12 at 7:54
If there was an elementary antiderivative, I'm sure Maple or Mathematica would find it. I'm pretty sure there isn't one. –  wj32 Nov 3 '12 at 7:57
Yeah, I just noticed that wouldn't come out right. That 1 as the lower limit was going to be a problem. –  Mike Nov 3 '12 at 9:19