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?

  • 1
    $\begingroup$ They are equal. Why didn't you link to the old question? $\endgroup$ – anon Mar 14 '12 at 6:39
  • 1
    $\begingroup$ The two answers are the same. $\endgroup$ – André Nicolas Mar 14 '12 at 6:41
  • $\begingroup$ @anon You'd have to dig around to find what I'm references. $\endgroup$ – Oliver Spryn Mar 14 '12 at 6:42
  • $\begingroup$ All the more reason to provide a link, I would say. -1 for the comment. $\endgroup$ – TonyK Jun 20 '19 at 11:39

I'm not sure what question you're referring to, so I can't address the misunderstanding, but

$$\int \cos(2x)dx=\frac{\sin(2x)}{2}+C=\sin(x)\cos(x)+C$$

is correct.

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  • 3
    $\begingroup$ That is because $sin(2x) = 2 sin(x) cos(x)$, which is what he needs to know explicitly $\endgroup$ – Jeremy Carlos Mar 14 '12 at 11:46

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