# Integration of a square of a derivative $\int f''(x)dx$

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.

• How about you work the other way and differentiate $\frac{1}{2}(c^2-1)(f')^2 - \cos f = a$ – Doug M Oct 20 '16 at 2:12
• The given form is supposed to evoke the chain rule, not integration by parts. – dxiv Oct 20 '16 at 2:12
• 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} – Sean Lake Oct 20 '16 at 2:13
• Oh right! So obvious now, thank you. – filterjuice Oct 20 '16 at 2:24

You have that $f'' f'$ is the derivative of $(f')^2/2$, and $(\sin f)f'$ is the derivative of $\cos f$.