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

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)".

in case i have mispelled the integral
it is integral of (1+ cos^2 (x) ) dx

 is it the same result?

You said sqrt(1+cos^2(x)) before.

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

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?

It's definitely sqrt(1 + cos² x), and not sqrt(1 - cos² x) ?
:)

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?

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?

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?

i know ...i read the article..but i am give then same formula ....

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)

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

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!

lol no problem :)