Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
They are equal. Why didn't you link to the old question? – anon Mar 14 '12 at 6:39
The two answers are the same. – André Nicolas Mar 14 '12 at 6:41
@anon You'd have to dig around to find what I'm references. – Oliver Spryn Mar 14 '12 at 6:42
Thank you all!! – Oliver Spryn Mar 14 '12 at 6:42
up vote 2 down vote accepted

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.

share|cite|improve this answer
That is because $sin(2x) = 2 sin(x) cos(x)$, which is what he needs to know explicitly – Jeremy Carlos Mar 14 '12 at 11:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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