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- #1

This integral arose when solving for a solitary wave in the KdV equation.

In class, the professor came up with:

\[

\int\frac{dx}{\sin(x)} = -\frac{1}{2}\ln\left[\frac{1 + \cos(x)}{1 - \cos(x)}\right]

\]

However, Wolfram obtains

\[

\int\frac{dx}{\sin(x)} = -\ln\left[\frac{1 + \cos(x)}{1 - \cos(x)}\right]

\]

but that changes our solitary waves solution from

\[

\sqrt{2\mu}e^{i\varepsilon^2\mu t}

\sech\left[\sqrt{\mu}(x - vt - x_0)\right]

\]

to

\[

\sqrt{2\mu}e^{i\varepsilon^2\mu t}

\sech\left[\frac{\sqrt{\mu}}{2}(x - vt - x_0)\right]

\]

For a solitary wave, the relation between the amplitude and the width has to at a certain portion. The professors solution for the solitary wave is correct but his integral doesn't workout compared to Wolframs.

Is there something I am missing?