1
$\begingroup$

We have $f(x) = (2x+4) \cos(2x)$

My method was integration by parts, and this is my calculation:

$$ \displaystyle \int \cos(2x) \cdot (2x+4) dx = \sin(2x)\cdot (2x-4) - \displaystyle\int \sin(2x) \cdot 2 dx$$

So our answer is:

$$ \displaystyle \int \displaystyle \cos(2x) \cdot (2x+4) dx = \sin(2x) \cdot (2x-4) + \cos(2x) +c$$

Apparently, this is incorrect; but I have no idea why. Can anyone point out my mistake?

$\endgroup$
  • 3
    $\begingroup$ The primitive of $\cos (2x)$ is $\frac12 \sin(2x)$. And you changed $(2x + 4)$ to $(2x - 4)$. $\endgroup$ – Daniel Fischer Jun 26 '13 at 15:42
  • $\begingroup$ @DanielFischer Answer it in the question so I can give you the best answer. I hate to ask these types of questions, but my mind is just weird; once I get a solution I can't seem to figure out what's wrong, if the solution is wrong. $\endgroup$ – ByeByeYa Jun 26 '13 at 15:46
  • $\begingroup$ @DanielFischer You may consider turning your comment into an answer, so that this question is more likely to be removed from the "UNANSWERED" tab. $\endgroup$ – user1551 Jun 26 '13 at 15:48
1
$\begingroup$

Two mistakes. One is probably a typo, you changed the $(2x + 4)$ factor of the integrand to $(2x - 4)$.

The other is that you didn't account for the factor $2$ in applying the inverse of the chain rule, the primitive of $\cos (2x)$ is $\frac12\sin(2x)$, and then the same once more, the primitive of $\sin(2x)$ is $-\frac12\cos(2x)$, so overall we obtain

$$\int \cos (2x)\cdot(2x+4)\,dx = \frac12\sin (2x)\cdot(2x+4) - \int \sin (2x)\,dx = (x+2)\sin (2x) + \frac12\cos (2x) + c.$$

$\endgroup$

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.