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integral of sqrt (1+cos^2 x) dx

Posted on 2006-11-05
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any ideas  :? whats the substitution here ?
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Question by:c_hockland
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14 Comments
 
LVL 22

Expert Comment

by:NovaDenizen
ID: 17877635
Wolfram integrator ( http://integrals.wolfram.com/index.jsp ) says:

sqrt(2)* E(x | 1/2)

where E is EllipticE, "the elliptic integral of the second kind" ( http://documents.wolfram.com/mathematica/functions/EllipticE/ ), which says E(k | m) is the integral of sqrt(1 - m sin^2 x)dx from 0 to k.  Not that helpful.

Seems like there is a big class of functions like f(x), when you try to automatically integrate them you get a result like "g(x), where g(x) is defined as the integral of f(x)".

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Author Comment

by:c_hockland
ID: 17877652
in case i have mispelled the integral
it is integral of (1+ cos^2 (x) ) dx

 is it the same result?
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LVL 22

Expert Comment

by:NovaDenizen
ID: 17877686
You said sqrt(1+cos^2(x)) before.

1+cos^2(x) is easy.  That's 1/4(6x + sin(2x))
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Author Comment

by:c_hockland
ID: 17877711
Nova Denizen , my apologies....seems i am so tired now....
the integral is  sqrt (1+ cos^2 (x)) dx

so i guess it is the first answer...(although for calc II problem seems kind of ...beyond the scope)  dont you think?
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LVL 25

Expert Comment

by:InteractiveMind
ID: 17877947
It's definitely sqrt(1 + cos² x), and not sqrt(1 - cos² x) ?
:)
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Author Comment

by:c_hockland
ID: 17877958
unfortunately it is  sqrt(1 + cos² x).....wish it was sqrt(1 - cos² x)!!!

cause i have f(x)=sinx

and i need the integral of ( sqrt (1+ ((fx)' )^2  ) dx  as i need to eval an approximation

(sinx)' = cos x

this is how i came up with this integral sqrt(1 + cos² x)

any ideas?
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LVL 25

Expert Comment

by:InteractiveMind
ID: 17877997
hmm

no luck so far..

But I'm determined to find a simpler solution than sqrt(2)*E(x | 1/2)... lol


Is the full problem you're solving:

f(x)=sin x
Find integral of sqrt(1 + f'(x)²)dx ?

Or, do you have to do anything earlier than that?
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Author Comment

by:c_hockland
ID: 17878016
well the question says that by using L = integral sqrt(1+ ((fX)' ^2) )  from [a,b]

and f(x)=sinx  [0,3]

and then it says  " write the appropriate integral that repsresent arc length foe rach function  
 f(x)=sinx  [0,3]

this is a common geometric shape - do u know the answer ??
-----------------end of question -----------------------------------------------------

now , i am not quite sure if i have to evaluate the integral or not....

".....this is a common geometric shape - do u know the answer ?? " 

what does the professor mean by that? does she need the solution or is she asking for something else?
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LVL 25

Accepted Solution

by:
InteractiveMind earned 2000 total points
ID: 17878045
ah; you want to find the arc length..

see here: http://en.wikipedia.org/wiki/Arc_length
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Author Comment

by:c_hockland
ID: 17878051
i know ...i read the article..but i am give then same formula ....


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LVL 25

Expert Comment

by:InteractiveMind
ID: 17878088
You've solved it?

I realise now that I've actually given that article a full read that it doesn't actually help too much :-\ sorry

Is it possible that the professor has made a slight mistake? ..should have used a - instead of a + perhaps?  (wishful thinking)
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LVL 25

Expert Comment

by:InteractiveMind
ID: 17878094
According to Dr Math, sqrt(1+cos²x) has no elementary antiderivative.
http://mathforum.org/library/drmath/view/52038.html
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Author Comment

by:c_hockland
ID: 17878112
yes, ....i told you tonight  i am very tired....

for this y=sinx the question states to just write the appropriate integral that represent the arc length

where it is required to provide solution (on the integral) it is written next to the question number..for the rest only the integral is required....

you definitely deserve the points for your time and patience with me tonight..

thanks mucho!

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LVL 25

Expert Comment

by:InteractiveMind
ID: 17878119
lol no problem :)
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