5
$\begingroup$

I took a shot in the dark and assumed that this is similar to solving $\int e^{x}\sin{x}\ dx$, but wolfram is giving me a different answer than what I got, and on top of that, I tried to differentiate my result and am not getting back what I started with. It's putting into question whether I was doing previous questions right or not..

First step of my attempt:

  • let $u=\cos(2x),\ du=-2\sin(2x)\ dx$
  • let $dv=\cos(3x)\ dx,\ v=\frac{\sin(3x)}{3}$

$$\int\cos(2x)\cos(3x)\ dx=\frac{\cos(2x)\sin(3x)}{3}+\frac{2}{3}\int\sin(2x)\sin(3x)\ dx $$

Then I did IBP again:

  • let $u=\sin(2x),\ du=2\cos(2x)\ dx$
  • let $dv=\sin(3x)\ dx, v=-\frac{cos(3x)}{3}$

$$=\frac{\cos(2x)\sin(3x)}{3}+\frac{2}{3}\left[-\frac{\cos(3x)\sin(2x)}{3}+\frac{2}{3}\int\cos(2x)\cos(3x)\ dx\right]$$

From there, I simplify and re-arrange to get

$$\frac{1}{3}\int\cos(2x)\cos(3x)\ dx=\frac{3\cos(2x)\sin(3x)-2\cos(3x)\sin(2x)}{9}$$ $$\int\cos(2x)\cos(3x)\ dx=\frac{3\cos(2x)\sin(3x)-2\cos(3x)\sin(2x)}{3}+C$$

So where did I go wrong? Wolfram says the answer should be

$$\int\cos(2x)\cos(3x)\ dx=\frac{1}{10}5\sin(x)+\sin(5x)+C$$

$\endgroup$
  • 2
    $\begingroup$ Don't integrate by parts. Use the appropriate produc to sum trigonometric formula: $\cos \theta\cos \phi=\ldots$. You can find it here: en.wikipedia.org/wiki/List_of_trigonometric_identities $\endgroup$ – Julien Feb 28 '13 at 18:06
  • $\begingroup$ I kind of like the idea of computing an integral in two different ways. You can get some interesting identities that way. Although the identity you get this way seems to essentially be a product to sum identity. $\endgroup$ – Baby Dragon Feb 28 '13 at 18:20
14
$\begingroup$

You have $\cos x \cos y = \frac{1}{2}(\cos (x+y) + \cos(x-y))$.

Hence $\cos (2x) \cos (3x) = \frac{1}{2} (\cos (5x) + \cos x) $. This should be straightforward to integrate.

$\endgroup$
  • 4
    $\begingroup$ Or use $\cos x = \frac{1}{2}(e^{ix}+e^{-ix})$. $\endgroup$ – copper.hat Feb 28 '13 at 18:08
2
$\begingroup$

You forgot to distribute the $\frac23$ after your second IBP. You should have $$\int\cos(2x)\cos(3x)\,dx=\frac{\cos(2x)\sin(3x)}3-\frac{2\cos(3x)\sin(2x)}9+\frac49\int\cos(2x)\cos(3x)\,dx.$$ From there, product-to-sum laws should get you the rest of the way (though it would be easier to simply use them from the start).

P.S.: Don't forget the integration constant!

$\endgroup$
  • $\begingroup$ Alternatively, you could subtract $\displaystyle \frac 49 \int \cos(2x) \cos(3x) dx$ from both sides and divide by $\displaystyle \frac 59$. $\endgroup$ – Joe Z. Feb 28 '13 at 18:40
  • $\begingroup$ @JoeZeng: Yes, but the OP knows to do that. The issue was that instead of $\frac49,$ the OP had $\frac23,$ so instead of $\frac59,$ the OP had $\frac13.$ $\endgroup$ – Cameron Buie Feb 28 '13 at 18:42
1
$\begingroup$

You did not go wrong, apart from a minor mistake in arithmetic. The denominator should be $5$.

However, the "product to sum" approach that Alpha took is, unusually for Alpha, more efficient.

Remark: You mention that you differentiated your final result and the derivative did not agree with the integrand. It is always possible to make a mistake in differentiating. When I do it, I get $\frac{5}{3}\cos 2x\cos 3x$. That says that your integral is almost right, and suggests that there was an unimportant glitch in the calculation. By the way, I computed the integral by using the Method of Undetermined Coefficients, looking for $A$ and $B$ such that $A\cos 2x\sin 3x +B\cos 3x\sin 2x$ works.

$\endgroup$
  • $\begingroup$ And, even if you do yours right, still Alpha's answer will look different. Even though they are both right answers. $\endgroup$ – GEdgar Feb 28 '13 at 18:27
  • $\begingroup$ OK - so I see I did do it right, except for missing the $\frac{2}{3}$ on the second IBP, thus messing up my answer enough to make my differentiation in the end incorrect. But I don't think the professor requires me to use the product-to-sum rules.. I used it anyhow, but thanks for pointing it out. $\endgroup$ – agent154 Feb 28 '13 at 18:31
  • $\begingroup$ You are welcome. You do the calculations well, and, importantly, give a thorough and well-formatted account of what you did. It would be unfortunate if a little slip made you lose faith in the soundness of your approach. $\endgroup$ – André Nicolas Feb 28 '13 at 18:44
  • $\begingroup$ Eternal vigilance is the price of correct computations... $\endgroup$ – copper.hat Feb 28 '13 at 19:28

protected by Zain Patel Sep 25 '17 at 11:14

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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