# Question about integration with inverse trigonometric functions

1. Dec 13, 2011

### pc2-brazil

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

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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:
$$\frac{s}{a} = \cos kt$$
$$s = a\cos kt$$

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 $\frac{s}{a} = \cos kt$.