Trig integral $\int ( \cos{x} + \sin{x}\cos{x}) \, dx $ Assume we have:
$$ \int (\cos{x} + \sin{x}\cos{x}) \, dx$$
Two ways to do it:
Use $$\sin{x}\cos{x} = \frac{ \sin{2x} }{2} $$
Then
$$ \int \left(\cos{x} + \frac{\sin{2x}}{2} \right) \, dx = \sin{x} - \frac{ \cos{2x} }{ 4 } + C. $$
The other way, just see that $ u = \sin(x), du = \cos(x)dx $,
$$ \int ( \cos{x} + \sin{x}\cos{x}) \, dx = \sin{x} + \frac{\sin^2{x}}{2} + C. $$
Now the part I don't see fully is, why aren't these results completely equal?
Taking the 2nd result,
\begin{align}
\sin{x} + \frac{\sin^2{x}}{2} &= \sin{x} + \frac{1}{2} \, \left( \frac{1 - \cos{2x}}{2} \right) \\
&= \sin{x} + \frac{1}{4} - \frac{\cos{2x}}{4}.
\end{align}
So you have to absorb $\frac{1}{4}$ into $C$ for them to be equal.
Shouldn't they be equal straight away?
 A: Your question boils down to this:
If $\int f(x) dx = F(x) + C$ and $\int f(x) dx = G(x) +C$ are both correct, then shouldn't it be true that $F(x)=G(x)$?
The answer is no.  $\int f(x)dx=F(x) +C$ means that (on the relevant interval, in this case all of $\mathbb{R}$) every antiderivative of $f(x)$ has the form $F(x)+C$ for some constant $C$.   The indefinite integral is really referring to a set of functions, namely all of the functions (on the relevant interval) whose derivatives equal $f(x)$.  If $F'(x)=f(x)$, then that set can be written as $\{F(x)+C:C\in \mathbb{R}\}$.  But the set of functions of the form $F(x)+C$ for some constant $C$ is precisely the same as the set of functions of the form $F(x)+22+C$ for some constant $C$, for instance.  Explicitly, $\{F(x)+C:C\in\mathbb{R}\}=\{F(x)+22+C:C\in\mathbb{R}\}$.  That is, if $F(x)$ and $G(x)$ only differ by a constant and $F(x)$ is an antiderivative of $f(x)$, then $\int f(x)dx=G(x)+C$ is also correct.
A: If you write 
$ \int{ \cos{x} + \frac{\sin{2x}}{2} dx }=\sin{x} - \frac{ cos{2x} }{ 4 } + C_1 $
and
$ \int{ \cos{x} + \sin{x}\cos{x} dx }= \sin{x} + \frac{\sin^2{x}}{2} + C_2 $, 
then you can see what's going on with the constants in your two cases($C_1$ and $C_2$). Being careful with choosing the notation is good for understanding.
