An antiderivative of an odd function is even. Proof in general. We have:


*

*$\int f(x)dx=F(x)+C$,

*$f(x) = -f(-x)$ - odd function.


Proof that $F(x)$ - even function.

My suggetion (please, check it): 


*

*Since 
$\int f(x)dx = F(x) + C$, 
$\int f(-x)dx = -F(-x) + C$.

*Since
$f(x)=-f(-x)$,
$\int f(x)dx=-\int f(-x)dx$

*Hence,
$F(x) + C = F(-x) + C \Rightarrow F(x)=F(-x)$

The nuance is that those constants are not necessarily the same. I mean, I know that this statement is correct, but I cant find the justification of this C-issue.
Even more. If eberything above is true, can we use the same technique to prove the reverse statement, i.e. an antiderivative of an even function is odd?
 A: Your idea is correct, but can be streamlined.
You can prove that two differentiable functions are equal by proving that they have the same derivative and the same value at some point. Equivalently, that their difference is constant (derivative zero) and has value zero at some point.
Let $f$ be continuous over $\mathbb{R}$ and odd. Let $F$ be an antiderivative of $f$. Then we can consider $G(x)=F(x)-F(-x)$. Then, using the chain rule,
$$
G'(x)=F'(x)+F'(-x)=f(x)+f(-x)=f(x)-f(x)=0
$$
and $G(0)=F(0)-F(0)=0$.
Can you prove that an antiderivative of an even function $f$ is odd? Not in general. There is only one antiderivative of an even function that's odd, precisely the only antiderivative $F$ such that $F(0)=0$.
A: Basically, in step $3$ you use the initial value for the odd function $f(-0)=-f(0)=0$:
$$f(-x)=-f(x) \iff \\
\int f(-x)dx = \int -f(x)dx \iff \\
-F(-x)+A=-F(x)+B \iff \\
\require{cancel}\cancel{-F(-0)}+A=\cancel{-F(0)}+B \iff \\
A=B.$$
A: If $f$ is odd,
$$F(x):=\int_0^x f(x)\,dx+C\implies
\\F(-x)=\int_0^{-x}f(x)\,dx+C=-\int_0^{x}f(-x)\,dx+C=F(x).$$
If $f$ is even,
$$F(x):=\int_0^x f(x)\,dx+C\implies F(0)=C$$ and in general $F$ is not odd.
