anonymous
  • anonymous
Help with integration. Can someone show me how to do this.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_0^{2 pi} (\sqrt{1/(3 pi)}+e^{i x}\sqrt{1/(6 pi)} )^2 dx = 2/3 \]
anonymous
  • anonymous
The inside is squared.
anonymous
  • anonymous
The e^ix is really bugging me.

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UnkleRhaukus
  • UnkleRhaukus
\[\int\limits_0^{2\pi}\sqrt{\frac{1}{3\pi}}+e^{ix}\sqrt{\frac{1}{{(6\pi)}^2}}\text d x=\frac 2 3\]
UnkleRhaukus
  • UnkleRhaukus
is that right?
anonymous
  • anonymous
no no The whole thing inside the integral is squared.
anonymous
  • anonymous
That's why I have a parentheses in the beginning it just doesn't show it that well when i used equations.
anonymous
  • anonymous
yes yes.
UnkleRhaukus
  • UnkleRhaukus
\[\int\limits_0^{2\pi}\left( {\sqrt{\frac{1}{3\pi}}+e^{ix}\sqrt{\frac{1}{6\pi}}}\right)^2\text d x=\frac 2 3\]
anonymous
  • anonymous
So do I just foil it out and find the integral?
inkyvoyd
  • inkyvoyd
turn e^(ix) into cos x+i sin x.
inkyvoyd
  • inkyvoyd
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.
inkyvoyd
  • inkyvoyd
Then, replace the g's back with i's. Remember that i^2=-1
anonymous
  • anonymous
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?
anonymous
  • anonymous
Can you copy and paste your integral it into Wolfram Alpha? It will show you how to do it step-by-step
anonymous
  • anonymous
tried that
inkyvoyd
  • inkyvoyd
I would input that into mathematica, but unfortunately my laptop with its installation is being fixed atm
anonymous
  • anonymous
My calc 2 teacher avoided imaginary numbers. So I'm really curious now.
anonymous
  • anonymous
Mathematica is on my dead mac... Ugh
anonymous
  • anonymous
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.
anonymous
  • anonymous
Anyone have Mac OS X up. Grapher could probably do it.
anonymous
  • anonymous
Change it to polar form maybe?
anonymous
  • anonymous
No the thing is that I was suppose to multiply imaginary function with -imaginary and that just cancels out all the imaginaries.
UnkleRhaukus
  • UnkleRhaukus
what do i enter into grapher? @Christbot
anonymous
  • anonymous
Wait, wait never about Grapher... Here http://en.wikipedia.org/wiki/Integration_using_Euler%27s_formula
anonymous
  • anonymous
You guys familiar with Euler's formula?
anonymous
  • anonymous
Yes but I don't need to use it at this point.
anonymous
  • anonymous
Have you tried foiling that giant square and breaking up the integral and placing the e^ix before the integral?
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=+integrate+from+0+to+2*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
anonymous
  • anonymous
That was the original problem I just wrote it wrong.
anonymous
  • anonymous
Ouch!
anonymous
  • anonymous
I actually got 1 with the addition of some leftover integrals that should cancel out yet I don't really know why.
anonymous
  • anonymous
Ouch indeed I hate my life right now. lol
anonymous
  • anonymous
shouldn't post your phone number on the internet lol and I don't have an iphone. :(
anonymous
  • anonymous
I'm just facebooking myself the input... I'll work to copy and paste, lol
anonymous
  • anonymous
What the hell?! Wolfram won't integrate it?!
anonymous
  • anonymous
Can't we let 1 = trig identity? Wow...
anonymous
  • anonymous
bam I got it as an indefinite integral! Gotta wait for my slow phone, sorry!
anonymous
  • anonymous
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anonymous
  • anonymous
TY!!
anonymous
  • anonymous
Want the last bit or you got it?
anonymous
  • anonymous
Go it!
anonymous
  • anonymous
High five!
anonymous
  • anonymous
*high five* thanks again!! really helped me out!
anonymous
  • anonymous
No prob! I love a good problem.
anonymous
  • anonymous
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 http://mathstem.blogspot.com/ I'll delete my cell# too, lol!

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