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I was trying to find the equation of a sinusoidal function that has one upward facing semicircle and immediately after one downwards facing semicircle which would make up a period of the function. When I evaluated the Fourier Series I came up with this integral:

$\int_{-1}^1 cos{(n\pi x)}\sqrt{1-x^2}~dx$

Also a clearer problem of similar difficulty is if the period is simply an upwards facing semicircle. If you can solve the integral or find another solution to the problem help is greatly appreciated.

user5200731
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    This produces a Bessel function. For example, see the derivation within this answer: http://math.stackexchange.com/questions/329347/how-to-calculate-int-01-arcsinx-sin-frac-pi2xdx/329355#329355 – Ron Gordon Feb 18 '16 at 02:47
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    This is a great comment. I did not know about this before! – Bombyx mori Feb 18 '16 at 02:53

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Using the parity of the integrand, in conjunction with a simple trigonometric substitution, will automatically yield a Bessel function, or, better said, the sum between a Bessel function and its second derivative, which can then be simplified using its various properties.

$$\begin{align} I~&=~2~\int_0^\tfrac\pi2\cos(n\pi\sin t)~\cos^2t~dt~=~2~\int_0^\tfrac\pi2\cos(n\pi\sin t)~\Big(1-\sin^2t\Big)~dt~=~ \\\\ ~&=~2~\int_0^\tfrac\pi2\cos(n\pi\sin t)~dt~+~2~\bigg[~\frac{\partial^2}{\partial a^2}\int_0^\tfrac\pi2\cos(a\sin t)~dt~\bigg]_{a=n\pi}~=~ \\\\ ~&=~\pi~\bigg[~J_0(n\pi)+J_0''(n\pi)~\bigg]~=~\frac\pi2~\bigg[~J_0(n\pi)+J_2(n\pi)~\bigg]~=~\frac{J_1(n\pi)}n~. \end{align}$$

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Lucian
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