Pleaes help solve this hardcore MATH problem!! HARD!!

[gwlam12]

The integral from 0 to 2pi of (cos(2x))/(5+4cosx) ...make sure you remmeer to change the x's to z's and that cos (2x) = 2(cos 2(x))^2 - 1. and remeber that cos z = (z^-1+z)/2. Use residues

 

[Ionizer86]

TI-89 🙂

Actually though, I think this may (or may not) be a case for partial fractions. I can't remeber if partial fractions fit though.

Which calc class are you in?

 

[gwlam12]

Book answer pi/6

Friend's answer 17pi/12

 

[Spencer278]

cos z = (z^-1+z)/2 Umm that isn't true (1^(-1) + 1 ) / 2 = 1 cos( 1 ) = .5403

 

[Rogue]

Advanced math is dumb. 😉

 

[Spencer278]

To start multiply by the inverse of (5+4cosx) over the inverse of (5+4cosx) so that your multipling by one. That will get ride of your nasty denominator.

 

[Rastus]

41, or is it 42?

It's one of them.

 

[dighn]

residues? is this complex variable analysis?

 

[gwlam12]

Yep

 

[her209]

The answer is

sqrt((3pi^23+72c)/[F*(V+d)])

 

[dighn]

i dont remember too much from that class heh.. 🙁


*adding it to my list of things to read

 

[RaynorWolfcastle]

Book is right, the answer is Pi/6. I just checked it myself

just use Euler's formula for the cosines then substitute exp(ix) = z and solve

 

[gwlam12]

nm

 

[RaynorWolfcastle]

apparently whoever took over your account is interested in complex variable math 😉

 

[gwlam12]

nm

 

[Muzzan]

Originally posted by: Spencer278
To start multiply by the inverse of (5+4cosx) over the inverse of (5+4cosx) so that your multipling by one. That will get ride of your nasty denominator.

I'm guessing you mean the conjugate, and not the inverse? Anyway, neither works here, since trigonometric functions are not square roots. 😉

 

[GermyBoy]

WTF are residues? I figured it out and had no problems doing it. Easy problem.

cos 2x = cos<SUP>2</SUP> x ? sin<SUP>2</SUP> x = 2 cos<SUP>2</SUP> x ? 1 = 1 ? 2 sin<SUP>2</SUP> x

This makes it stupidly easy.

 

[miniMUNCH]

Originally posted by: GermyBoy
WTF are residues? I figured it out and had no problems doing it. Easy problem.

cos 2x = cos<SUP>2</SUP> x ? sin<SUP>2</SUP> x = 2 cos<SUP>2</SUP> x ? 1 = 1 ? 2 sin<SUP>2</SUP> x

This makes it stupidly easy.

Residues are something you would learn in complex variable calculus...and therefore, since he is supposed to solve the problem using the Residue Theorem, the problem isn't stupidly easy now is it.

So don't be such a retard next time.

 

[RaynorWolfcastle]

Originally posted by: GermyBoy
WTF are residues? I figured it out and had no problems doing it. Easy problem.

cos 2x = cos<SUP>2</SUP> x ? sin<SUP>2</SUP> x = 2 cos<SUP>2</SUP> x ? 1 = 1 ? 2 sin<SUP>2</SUP> x

This makes it stupidly easy.

did you notice that cos in the denominator? Please, try your method and if it works I'd like to hear it.

 

[vtqanh]

Originally posted by: RaynorWolfcastle

Originally posted by: GermyBoy
WTF are residues? I figured it out and had no problems doing it. Easy problem.

cos 2x = cos<SUP>2</SUP> x ? sin<SUP>2</SUP> x = 2 cos<SUP>2</SUP> x ? 1 = 1 ? 2 sin<SUP>2</SUP> x

This makes it stupidly easy.

did you notice that cos in the denominator? Please, try your method and if it works I'd like to hear it.

I'd like to know, too.

 

[yoda291]

according to math league and all my advanced mathematics course notes....

the answer is either 0,1,-1, infinity, negative infinity, or undefined.

pick one.