2
$\begingroup$

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?

$\endgroup$
  • 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
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.