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I was reading a textbook on PDEs and Waves, and it had a line of integration which did not make any sense to me:

$$(c^2 - 1)f''f'+(\sin f)f' =0 $$ can be integrated to produce the first order equation: $$ \frac{1}{2}(c^2-1)(f')^2 - \cos f = a$$

I tried integration by parts which gave me $$(c^2-1) \left((f')^2 - \int(f'')^2\right) - f' \cos f + \int f'' \cos f$$ and now I'm stuck, and not sure how to proceed. I'm not sure how to integrate $\int (f'')^2$, and how to get rid of the factor in front of the trig function.

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    $\begingroup$ How about you work the other way and differentiate $\frac{1}{2}(c^2-1)(f')^2 - \cos f = a$ $\endgroup$ – Doug M Oct 20 '16 at 2:12
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    $\begingroup$ The given form is supposed to evoke the chain rule, not integration by parts. $\endgroup$ – dxiv Oct 20 '16 at 2:12
  • $\begingroup$ hints:$$\begin{align} \frac{\operatorname{d}}{\operatorname{d}t} \cos(f(t)) &= -\sin f f' \\ \frac{\operatorname{d}}{\operatorname{d}t} (f')^2 & = 2f' f''\end{align}$$ $\endgroup$ – Sean Lake Oct 20 '16 at 2:13
  • $\begingroup$ Oh right! So obvious now, thank you. $\endgroup$ – filterjuice Oct 20 '16 at 2:24
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You have that $f'' f'$ is the derivative of $(f')^2/2$, and $(\sin f)f'$ is the derivative of $\cos f$.

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