How to reconcile the fact that the antiderivative of $\sin(x)\cos(x)$ has two possible answers? So here is the problem. 
$$
\int\sin(x)\cos(x)dx = \frac{1}{2}\sin^2(x) + c_1 \\
= -\frac{1}{2}\cos^2(x) + c_2
$$
This fact doesn't make much sense to me. How do you reconcile the fact that there are two possibilities (notice that these answers aren't only different by a constant)? 
What are some other functions that have more than one antiderivatives (not only different by a constant)? 
 A: Note:
$$
\begin{split} \frac{1}{2}\sin^2(x)+c_1 &= \frac{1}{2}(1-\cos^2(x))+c_1 \\ &=\frac{1}{2}-\frac{1}{2}\cos^2(x)+c_1 \\ &=-\frac{1}{2}\cos^2(x)+c_2  \end{split}
$$
where $c_2=c_1+\frac{1}{2}$. The answers are both the same so there is no contradiction.
A: There is no contradiction: it illustrates the fact that trigonometry formulae are polymorphic.
You even have a third answer:
$$-\frac14 \cos2x+\text{constant}.$$
A: You wrote:

(notice that these answers aren't only different by a constant)

But they do differ by a constant.  If you subtract one solution from the other, you have
$$ ( \frac{1}{2}\sin^2(x)+c_1 ) - ( -\frac{1}{2}\cos^2(x)+c_2 )$$
which is equal to
$$\frac{1}{2} (\sin^2(x) + \cos^2(x) ) + c_1 - c_2$$
and using the trig identity $\sin^2(x) + \cos^2(x) = 1$, all of this is equal to just
$$\frac{1}{2} + c_1 - c_2$$
which is a constant.
A: Since $\ \sin^2(x)+\cos^2(x)=1$, if we let $c_1=c_2-\frac 12$ so we'd have
$$\begin{align}
\frac{1}{2}\sin^2(x) + c_1 &= \frac{1}{2}\sin^2(x) + (c_2-\frac 12) \\
&= \frac{1}{2}\sin^2(x) + (c_2-\frac 12) \\
&= \frac{1}{2}(1-\cos^2(x)) + (c_2-\frac 12) \\
&= c_2-\frac{1}{2}\cos^2(x)
\end{align}$$
