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As stated the title, I get to a point which I can't do anything, and I'm sure I've made a mistake some where, here is my full working out:

$$ \int e^{ix}\cos(x)dx \\ u = e^{ix} \text{ | } u'= ie^{ie} \\ v = \sin(x) \text{ | } v'=\cos(x) \\ e^{ix}\sin(x)-\int ie^{ix}\sin(x)dx + C \\ e^{ix}\sin(x)-i\int e^{ix}\sin(x)dx + C \\ u = e^{ix} \text{ | } u'= ie^{ie} \\ v = -\cos(x) \text{ | } v'=\sin(x) \\ e^{ix}\sin(x)-i(-e^{ix} \cos(x) + i \int e^{ix}cos(x)dx) + C \\ Let \int e^{ix}\cos(x)dx = I \\ I = e^{ix}(\sin(x) + i\cos(x)) - i^2I + C $$

But ($i^2 = -1$) so the equation should become: $$ I = e^{ix}(\sin(x) + i\cos(x)) + I + C $$ And this is where I'm stuck, I can't simply take $I$ away from both sides, that would make $$e^{ix}(\sin(x) + i\cos(x)) = 0$$ What have a messed up in the process? And just to make it clear someone in a previous question didn't under understand what $v'$ was, it's the same as $\frac{dv}{dx}$, thank you in advance.

Sam Chahine
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  • Related : http://math.stackexchange.com/questions/540295/integrate-eax-sinbx – lab bhattacharjee Jan 16 '14 at 04:28
  • I already read that question, but unfortunately didn't help me, I think the existence of the imaginary number $i$ in my question puts it in a seperate category than the one you linked, I might be wrong, but I did use integration by parts, thank you! – Sam Chahine Jan 16 '14 at 04:32
  • What a stupid mistake! Thank you, I should pay more attension next time, sorry for the trouble, thank you again. – Sam Chahine Jan 16 '14 at 04:37
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    You can also use $e^{ix}=\cos x+i\sin x$, which simplifies your calculations. – daulomb Jan 16 '14 at 04:40
  • Very interesting, thank you everyone! – Sam Chahine Jan 16 '14 at 04:41
  • $e^{ix}(\sin(x)+i\cos(x))=e^{ix}(ie^{-ix})=i$. So the equation you got is $I=i+I+C$, which is correct for $C=-i$, but doesn't produce the value for $I$. The fact that you don't get the value of $I$ by this integration by parts is the same as why you don't get it for $\int\cos(x)\cos(x)dx$ and for $\int\sin(x)\cos(x)dx$. For these we can use the formulas for $\sin(2x)$ and $\cos(2x)$, which is equivalent to what user44197 did in his answer. – Pp.. Jan 09 '15 at 11:05

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$$\int e^{ix}\cos x\,dx=\int e^{ix}\frac{e^{ix}+e^{-ix}}{2}\,dx = \frac12 \int \left(e^{2ix} + 1\right) dx = \frac1 {4i} e^{2ix} + \frac x 2 $$

user44197
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  • This is simple and look correct. Thus, either the OP made a mistake before the point from where I took over, or else I did...or something. – DonAntonio Jan 16 '14 at 05:10
  • must be... I didn't check OP's first step. I just differentiated the right and side and I get the left hand side. – user44197 Jan 16 '14 at 05:11
  • This is the way I'd have done it to begin with, yet for some reason the OP chose by parts and stuff. I supposed he hasn't yet studied the complex form of the trigonometric function, or whatever. – DonAntonio Jan 16 '14 at 05:12
  • You're right, I discovered imaginary numbers a couple of days ago, the interested me so I just tended to implement them into problems I make up on a regular basis, where can I study that? And what's a good website to get familiar will all the trigonometric functions? Thank you – Sam Chahine Jan 16 '14 at 05:17
  • I forgot to tag you and I can't change my comment after 5 minutes, @DonAntonio, thank you – Sam Chahine Jan 16 '14 at 05:23
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    I already read it, @SamirChahine . Thanks. There are thousands of internet sites dealing with this stuff. Just google "Complex numbers" , and for trigonometric functions google...well, those two words. – DonAntonio Jan 16 '14 at 05:25
  • I'l do that right away, sorry for the hassle, thank you again! – Sam Chahine Jan 16 '14 at 05:26