Question about integration with inverse trigonometric functions

[pc2-brazil]

I'm self-studying Calculus and would like to ask some doubts about the following question:

Homework Statement

If, in t seconds, s is the oriented distance of the particle from the origin and v is the velocity of the particle, then a differential equation for harmonic simple motion is:
\frac{ds}{dt}=-k^2s
where k² is a proportionality constant.
Since \frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds},
v\frac{dv}{ds}=-k^2s
a) Obtain that v = \pm\sqrt{a^2-s^2}. Note: Take a²k² as the arbitrary integration constant and justify this choice.
b) Taking t = 0 in the instant that v = 0 (s = a), obtain that s = a cos kt.

Homework Equations

The Attempt at a Solution

a)
\int vdv=\int -k^2sds
\frac{v^2}{2}=-\frac{k^2s^2}{2}+\text{constant}
v^2=-k^2s^2+C
v=\pm\sqrt{C-k^2s^2}
v=\pm\sqrt{a^2k^2-k^2s^2}
Using C = a²k²:
v=\pm k\sqrt{a^2-s^2}
I'm not sure how to justify the choice of C as a²k². Is it because the velocity will be 0 for s = a (the amplitude of the motion)?
b)
\frac{ds}{dt}=\pm k\sqrt{a^2-s^2}
\frac{ds}{\sqrt{a^2-s^2}}=\pm k dt
\arcsin\frac{s}{a} = \pm kt + \bar{c} where a > 0
\frac{\pi}{2} - \arccos\frac{s}{a} = \pm kt + \bar{k}
-\arccos\frac{s}{a} = \pm kt + \bar{k} - \frac{\pi}{2}
Replacing \bar{k} - \frac{\pi}{2} by \bar{c}:
-\arccos\frac{s}{a} = \pm kt + \bar{c}
\arccos\frac{s}{a} = \mp kt - \bar{c}
When t = 0, s = a, so:
\arccos\frac{a}{a} = -\bar{c}
\arccos 1 = -\bar{c}
\bar{c}=0
Then:
\arccos\frac{s}{a} = \pm kt
Now the problem is that I'm not sure how to get rid of the ±.
I think that ± is eliminated here because, in the way that the arc-cosine function is defined, its image is from 0 to π (therefore, its image must be positive). Then:
\frac{s}{a} = \cos kt
s = a\cos kt

Thank you in advance.

 

[ehild]

Now the problem is that I'm not sure how to get rid of the ±.
I think that ± is eliminated here because, in the way that the arc-cosine function is defined, its image is from 0 to π (therefore, its image must be positive). Then:
\frac{s}{a} = \cos kt
s = a\cos kt

Thank you in advance.

Cosine is an even function, cos(x)=cos(-x). When you take the cosine of both sides of \arccos\frac{s}{a} = \pm kt it gives the single result \frac{s}{a} = \cos kt.


ehild