An interesting proof that $\sin^2(x) + \cos^2(x) = 1$ (using only series, no trigonometry). This question concerns an interesting proof of the fact that $\sin^2(x) + \cos^2(x) = 1$, but only using the series that defines them, not any trigonometry. So define
$$
s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots
$$
and 
$$
c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots
$$
Step 1: we prove that $s' = c$ and $c' = -s$. This can be done by differentiating the series componentwise:
$$
s'(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots = c(x), 
$$
and 
$$
c'(x) = - x + \frac{x^3}{3!} - \frac{x^5}{5!} + \ldots = -s(x).
$$
Step 2: we prove that $(s^2+c^2)' = 0$. Using the chain-rule on both terms and then using our result of step 1 we compute:
$$
(s^2+c^2)' = 2s \cdot s' + 2c \cdot c' = 2sc+2c(-s) = 0. 
$$
Step 3: we prove that $s^2 + c^2 = 1$. The idea here is to use step 2, to obtain something like 
$$
s^2 + c^2 = \int (s^2 + c^2)'\,dx = \int 0 \, dx = 1. 
$$
However, I cannot figure out the details of this last step. In particular, as far as I know, $\int (s^2 + c^2)'\,dx = s^2+c^2 + C$, and $\int 0\,dx = C'$. What happens with these constants?

Your last step is unnecessarily complicated.  In Step 2, you show that
$$ (s^2 + c^2)' = 0 \implies (s^2 + c^2)(x) = C, $$
where $C$ is some constant.  In particular,
$$ (s^2 + c^2)(0) = C. $$
But, directly from the power series definitions of $s$ and $c$, we have
$$ s(0) = \frac{0}{1!} - \frac{0^3}{3!} + \frac{0^5}{5!} + \dotsb = 0
\quad\text{and}\quad
c(0) = 1 - \frac{0^2}{2!} + \frac{0^4}{4!} + \dotsb = 1.$$
Therefore
$$ (s^2 + c^2)(x)
= (s^2 + c^2)(0)
= s(0)^2 + c(0)^2
= 0^2 + 1^2
= 1. $$
