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I am given $\frac{dv}{ds} = \sec(s-v)$

I have tried to solve this by using the substitution $t=\tan(\frac{\omega}{2})$,

but I am not sure how to proceed so that I can solve the equation such that $s = s(v)$.

Any help would be greatly appreciated!

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$$\frac{dv}{ds} = \sec(s-v)$$ As Igor proposed substitute $z=s-v \implies v'=1-z'$ $$ 1-z'=\sec(z)$$ $$z'=1-\sec(z)$$ $$ \int \frac {\cos(z)}{\cos(z)-1}dz=\int ds=s+K$$ $$z+ \int \frac {dz}{\cos(z)-1}=s+K$$ Note that $${\cos(z)-1}={\cos(z)-\cos(0)}=-2\sin^2(z/2)$$ $$z- \frac 12 \int \frac {dz}{\sin^2(z/2)}=s+K$$ $$z- \int \frac {du}{\sin^2(u)}=s+K$$ $$\boxed{v- \cot((s-v)/2)=K}$$

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First, note that

$$\frac{ds}{dv} = \cos(s-v).$$

Secondly, substitute $w = s - v,$ so $s = v + w,$ to get $$\frac{dw}{dv} + 1 = \cos{w}.$$

The rest should be reasonably self-explanatory.

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