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primeralph
 one year ago
Best ResponseYou've already chosen the best response.0I'll be honest, I can solve that......the solution is not neat......so don't expect anyone here to be able to finish it....it's long...........try wolfram....... but you still can try here though

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0can you tell me how to at least start it?

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0I feel like this will end up being an ln function, but I think there's a lot more to it than just that I'm sure. Let Ralph help you.

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0does anyone know of a completed example of a problem like this I can go online and review or anything?

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0dw:1368859626389:dw

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0That's how you start.

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0can you tell me what the J is for? I have never ever seen that used before

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0dw:1368859717310:dw

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0J is just a multiplier. In this case, J = 1 , so you can remove it

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0I highly doubt that it would be that simple felavin.

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0could the next step be to let u=tan(theta) then du=sec^2(theta) d(theta)

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@SinginDaCalc2Blues same thing.........you're looping

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0Yeah all that would do is put you right back to where you started

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0^^^ that girl is just too smart..........

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0I'm just here to learn how to do it myself

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0are you people calling me stupid? seriously?

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0No, I was calling myself stupid xD

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0no, I didn't mean that.

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0okay, let's figure out an easier way to solve this problem.

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@SinginDaCalc2Blues can you handle recursive integrals?

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0okay...well I appreciate the help from everyone but it's hard to understand the context of nonverbal communication sometimes...I was hoping you guys weren't calling me stupid...anyway...

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@SinginDaCalc2Blues we really would never do that.

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0i have never heard the term recursive integral...while I probably know what your talking about, I don't know it in those terms...can you describe?

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0wait............a minute...why not just polar?

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0Polar coordinates? Actually that makes some sense...

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0are you saying turn the integral into a polar expression or something?

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@smokeydabear tricky thing is, it returns the integrand to the form we had before....because using cos, cos = 1/sec.....

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0my professor is devious!

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0Your professor is pretty evil for giving you something like this

Peter14
 one year ago
Best ResponseYou've already chosen the best response.0would u=x^2 du=2x dx indefinite integral of 1/(u^4+1) help? (sorry, my latex isn't working)

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@SinginDaCalc2Blues I reduced it to......can you solve this?dw:1368860855998:dw

felavin
 one year ago
Best ResponseYou've already chosen the best response.1the int, called complex in

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0uhm if k is a constant, what is J again?

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0J is the variable of integration, just like U

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0U in U substitution

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0I just looked it up on wolfram... what is this monstrosity...

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@smokeydabear I told you...........

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0How do you know where to even start with something like this? That's always been my problem

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0a good place to start is with Vodka :o)

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@SinginDaCalc2Blues I asked you if you knew recursives because as you can see here, the trigs are repeating..which means recursion

Peter14
 one year ago
Best ResponseYou've already chosen the best response.0no, not vodka, start with despair.

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0in 30 minutes i will no longer be able to do integration but I will be able to slurrrr Russian! :o)~

Peter14
 one year ago
Best ResponseYou've already chosen the best response.0vodka inhibits your ability to do math.

Peter14
 one year ago
Best ResponseYou've already chosen the best response.0despair only temporarily inhibits your ability to do everything

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@smokeydabear recursion works to start. for exampledw:1368861204284:dw

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so how would we use that to help us? I'm still confused :s

some_someone
 one year ago
Best ResponseYou've already chosen the best response.2well long only :P

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@smokeydabear it saves time because once you know that it will repeat, you can generalize..........

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0hey guys...i have an idea...gimme a sec...

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0.........meaning............dw:1368861457508:dw

stamp
 one year ago
Best ResponseYou've already chosen the best response.0u = x^2, du = 2x dx\[\int \frac{du}{u^4+1}\]

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0so, because they occur in trig functions, we can use it for the tan theta

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0Ohhhhhh I see what you did with the sin's and the pi's @primeralph very clever! =D

Peter14
 one year ago
Best ResponseYou've already chosen the best response.0now we just need to learn to use it effectively...

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0dw:1368861979382:dw

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0I gotcha Ralph, I noticed it too but I didn't want to be mean =P

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0Does this little bit of trickyness open any doors?

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0yeah what stamp wrote I believe is wrong...it wouldn't be x^4...it would be (x^4)^2

smokeydabear
 one year ago
Best ResponseYou've already chosen the best response.0Well what stamp wrote isn't technically wrong, but it isn't the method to solving this.

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0I basically factored out of the denominator an x^2...then did usub

felavin
 one year ago
Best ResponseYou've already chosen the best response.1hi, the integrate is simple: Who can solve this integrate:\[\int\limits_{}^{} \frac{\tan Z }{ Z \frac{ P }{ 4}}\] GAME IS OVER ..........

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0hi...i'm only in calc 2...i don't know what CauchyReimann equations are... I did however get my integral down to 1/(u^3+1)

drawar
 one year ago
Best ResponseYou've already chosen the best response.0In case you love twisting with partial fractions: \[\frac{ 2x }{ x^8 +1}=\frac{ 2x }{( x^4\sqrt{2}x^2+1)( x^4+\sqrt{2}x^2+1)}\]

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0hi drawer...i think your way is looking promising...i think i was headed toward that...can you look at my .bmp and tell me how to start partial fraction with what i ended with?

drawar
 one year ago
Best ResponseYou've already chosen the best response.0Then \[u^3+1=(u+1)(u^2u+1)\], use partial fractions.

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0what do i do with the numerator?

drawar
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ u^3+1 }=\frac{ A }{ u+1 }+\frac{ Bu+C }{ u^2u+1 }\] =>\[A(u^2u+1)+(Bu+C)(u+1)=1\] Then equate the coefficients and you'll get a system of linear equations in terms of A,B, and C. Solve it!

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0so then...after I do that, will i have technically several different integrals which will be easier to integrate?

drawar
 one year ago
Best ResponseYou've already chosen the best response.0Yes, you'll see that the degree of the polynomial of the numerator is always 1 less than that of the denominator, which must have sth to do with the natural logarithm (ln) function.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0i havent read much what is written above, its too long, has the question been solved ?

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0ok drawar...the answer definitely has some nasty ln's...i will keep plugging away at it...one more question...

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0once i figure out what A, B, AND C equate to, don't I just plug those into the numerators and then treat one as a separate integral to solve?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0has anyone tried x^2 =t ?

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0yeah shub...wayyy up at the top...we are past that now...but thanks

drawar
 one year ago
Best ResponseYou've already chosen the best response.0Yes @SinginDaCalc2Blues

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0well, why you didnt go forward with that method, after that substitution, we'll have dt/(t^4 +1) note that t^4 + 1 = (t^2 + i) (t^2 i) also note that (t^2 +i)  ( t^ i) = 2i , a constant

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0ok...i will keep plugging away and see if I can get it...this problem has been so crazy! :o)

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0are you using imaginary numbers shub?

drawar
 one year ago
Best ResponseYou've already chosen the best response.0is it a definite/indefinite integral in your original question?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0yes, the very same. i can be treated as just another constant

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0i have no idea how to use imaginary numbers...i'm only in calc 2..we haven't used those

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0I see, well use of iota will lead you to the solution withing 2 more steps, and in the end you can always replace the complex part with hyperbolic functions, or natural logarithms . But anyways, since you havent learnt about that, go with @drawar 's method (:

SinginDaCalc2Blues
 one year ago
Best ResponseYou've already chosen the best response.0I wish I had learned about them...i just hope i can solve this

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1\[\int\frac{2x}{x^8+1}dx\]\[=\int \frac{1}{x^8+1}d(x^2)\]\[=\int \frac{1}{y^4+1}dy\]\[=\frac{1}{2}\int \frac{y^2+1y^2 +1}{y^4+1} dy\]\[=\frac{1}{2}(\int \frac{y^2+1}{y^4+1}dy\int\frac{y^2 1}{y^4+1} dy)\]\[=\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\int\frac{1 \frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)\] Consider \[\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\] Let u = \(y\frac{1}{y}\) \(du =(1 + \frac{1}{y^2}) dy\) and \(y^2 + 1/y^2 = (y\frac{1}{y})^2+2=u^2+2\) So, \[\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\]\[=\int \frac{1}{u^2+2}du\] Similarly, for\(\int \frac{1\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\) Let \(v = y+ \frac{1}{y}\) \[dv = 1  \frac{1}{y^2}dy\]\[y^2 + 1/y^2 = (y+\frac{1}{y})^22=v^22\] So, \[\int \frac{1\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\]\[=\int\frac{1}{v^22}dv\] As a result, \[\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\int\frac{1 \frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)\]\[=\frac{1}{2}(\int\frac{1}{u^2+2}du\int\frac{1}{v^22}dv)\]And You can integrate it then.

primeralph
 one year ago
Best ResponseYou've already chosen the best response.0@roly Thank you for taking the time to illustrate it.
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