For some reason, I am lost on one part of this integration problem:
$\begin{align}\int \cos{\left(2x\right)} \ dx\end{align}$
$u = 2x$
$du = 2 \ dx$
$\frac{1}{2} du = dx$
$\begin{align}\frac{1}{2} \int \cos{\left(u\right)} \ du\end{align}$
$\begin{align}\frac{\sin{u}}{2} + C\end{align}$
$\begin{align}\frac{\sin{\left(2x\right)}}{2} + C\end{align}$
The only issue is that I have been told that this should really equal $\sin{x}\cos{x} + C$. I know that $\sin{\left(2x\right)} = 2\sin{x}\cos{x} + C$, so:
$\begin{align}\frac{\sin{\left(2x\right)}}{2} = \frac{2\sin{x}\cos{x}}{2} + C = \sin{x}\cos{x} + C\end{align}$
On a question I had asked previously, I was told that $\begin{align} \int \cos{\left(2x\right)} \ dx \neq \frac{\sin{\left(2x\right)}}{2} + C\end{align}$, but rather $\sin{x}\cos{x} + C$.
Please excuse the silly question, but these two expressions are the same, right?