\(\[\int cost^2dt\]\) = ?

Thanks.
MichealY.

 

MichealY,

What are you trying to integrate? cos^2 (t)dt, cos(t^2), or what? Let's use some parentheses so we know what you what you are talking about. Can't you look that up in a table of integrals?

Ratch

 

Ratch:
It is cos(t^2) not cos^2(t).
No I cannot look up in a table of integrals.
I want to know how to solve it by us,altough it could be solved by MATLAB/Mathematica.

MichealY

 

Change variables from t^2 to x, then integrate by parts?

 

Nukular:
What do you mean by integrating by parts?

MichealY

 

Presumably you realize the solution is an infinite series.

 

t n k,
Bingo.I asked someone of mathematic major the other day.He said it is a problem of infinite series,But I do not even know what's infinite series,so I need the help of you,and at the same time I will work hard to learn infinite series

MichealY

 

Hi MichealY,

As Nukular rightly points out the solution probably requires an understanding of both substitution of variables and integration by parts. It's the approach I would use anyway. Your earlier post suggests you are not familiar with the latter method. It would be useful to do some background study on that first and then return to this problem. Are you studying integration as part of formal course or is this something simply of personal interest?

 

Nukular,
I have understood a bit of what you mean.

t n k,
Yes, I am not familiar with integration by parts,maybe I just don't know this English phrase I will do some background study and then return to this problem.I'm studying integration as part of formal course(I failed the final exam,now trying to make up),and it is something of personal interest too.

Thanks to both of you.

MichealY

 

Here is integration by parts in mathematics notation:
\(\int u dv = uv - \int v du\)

It is the product rule of differentiation in reverse.
The trick to using this technique is to identify what part of the original integrand is u, which the rest being called dv. From this you antidifferentiate dv to get v, and take the differential of u to get du. If you have chosen u and dv correctly, the new integral will be easier to find than the one you started with.

 

Mark44,
Thanks.I have had learnt integration by parts before.I thought it means,first,reduce cost^2 to many parts,then integrate them by parts.

I have also tried this method,but it seems that it won't work.

MichealY

 

I have also tried this method,but it seems that it won't work.

MichealY

It does work.

 

Recognizing that the solution is an infinite series, the other approach would be to expand cos(t^2) as an infinite series and integrate that wrt t. The trouble with any solution one obtains using any approach is getting the answer into an agreed "standard form" - i.e. as agreed by mathematicians. I haven't looked up a standard handbook of integral tables for this function, but I would think the agreed solution has long been cast in stone. Finding a derivation might be more difficult - I guess it's out there somewhere!

One could probably better spend time doing practice on any number of functions with finite solutions of increasing degree of difficulty.

Then return to this at leisure........if one has such luxury.

 

Of course you can always try wiki ....

http://en.wikipedia.org/wiki/Fresnel_integral

 

Here is the derivation, as a link to save me a lot of typing.

http://wiki.answers.com/Q/What_is_the_integral_of_sin_x_squared

 

Good find Studiot! - hopefully MichealY will find this helpful.

 

Thanks,you are really kind.

I have also tried this method

I mean I have tried method of integrating by parts.

And by using MATLAB,the result returned contain Fersnel C.

MichealY

 

Hi I also do think this is related to infinite series
try and use sandwicth theorem to find whether it is convergent and find it's limit,if i'm not mistaken this limit is the value of the integral

 

Starting to study numerical analysis,I have learnt a bit
It seems that langarage polynomail is better than taylor polynomial.