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Help with integration. Can someone show me how to do this.

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\[\int\limits_0^{2 pi} (\sqrt{1/(3 pi)}+e^{i x}\sqrt{1/(6 pi)} )^2 dx = 2/3 \]
The inside is squared.
The e^ix is really bugging me.

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Other answers:

\[\int\limits_0^{2\pi}\sqrt{\frac{1}{3\pi}}+e^{ix}\sqrt{\frac{1}{{(6\pi)}^2}}\text d x=\frac 2 3\]
is that right?
no no The whole thing inside the integral is squared.
That's why I have a parentheses in the beginning it just doesn't show it that well when i used equations.
yes yes.
\[\int\limits_0^{2\pi}\left( {\sqrt{\frac{1}{3\pi}}+e^{ix}\sqrt{\frac{1}{6\pi}}}\right)^2\text d x=\frac 2 3\]
So do I just foil it out and find the integral?
turn e^(ix) into cos x+i sin x.
If it really bothers you that much. I would just treat i as a constnat. Replace i with g and just integrate it like a constant.
Then, replace the g's back with i's. Remember that i^2=-1
If I carry out the foil I know that I can split it into three different integral with the first one being integrating 1/3pi If I do that I can simply take out 1/3pi since it's a constant and I get (1/3pi)(2pi-0)=2/3 which is the answer but how does that other part cancel out?
Can you copy and paste your integral it into Wolfram Alpha? It will show you how to do it step-by-step
tried that
I would input that into mathematica, but unfortunately my laptop with its installation is being fixed atm
My calc 2 teacher avoided imaginary numbers. So I'm really curious now.
Mathematica is on my dead mac... Ugh
Oh I think I know what I did wrong. I was suppose to multiply the inside function with it's complex conjugate so it cancels out all imaginar values.
Anyone have Mac OS X up. Grapher could probably do it.
Change it to polar form maybe?
No the thing is that I was suppose to multiply imaginary function with -imaginary and that just cancels out all the imaginaries.
what do i enter into grapher? @Christbot
Wait, wait never about Grapher... Here
You guys familiar with Euler's formula?
Yes but I don't need to use it at this point.
Have you tried foiling that giant square and breaking up the integral and placing the e^ix before the integral?*pi+%28%28%28squareroot%281%2F%283*pi%29%29%2Bsquareroot%281%2F%286*pi%29%29+e%5Eix%29%29*%28%28squareroot%281%2F%283*pi%29%29%2Bsquareroot%281%2F%286*pi%29%29+e%5E-ix%29%29%29
That was the original problem I just wrote it wrong.
I actually got 1 with the addition of some leftover integrals that should cancel out yet I don't really know why.
Ouch indeed I hate my life right now. lol
shouldn't post your phone number on the internet lol and I don't have an iphone. :(
I'm just facebooking myself the input... I'll work to copy and paste, lol
What the hell?! Wolfram won't integrate it?!
Can't we let 1 = trig identity? Wow...
bam I got it as an indefinite integral! Gotta wait for my slow phone, sorry!
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Go it!
High five!
*high five* thanks again!! really helped me out!
No prob! I love a good problem.
Oh the web version works for free. Yeah, sometime you just gotta tinker with the input. I'm deleting the image links, but here is my mathy blog, FWIW I'll delete my cell# too, lol!

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