
#1
Dec2804, 11:02 AM

P: 104

Hi, I just don't get this. I'm to lazy to type inn what I've done, so I just took
a picture of my textbook: http://home.no.net/erfr1/images/1.jpg http://home.no.net/erfr1/images/2.jpg You're supposed to end up with the starting point, right? So you can divide the rest by two? Whatever I try, I get zero... I got the solution from mathematica, but don't understand a thing of it. Thanks! 



#2
Dec2804, 11:16 AM

P: 696

I'm not exactly clear on what it is you have done, but I'm guessing that you tried to integrate cos^2(x) using partial integration, and the equation you got reduced to 0 = 0? I suppose you expected to get back your original integral after a few iterations, so that you could solve for it. I wouldn't say that's what's "supposed" to happen (but it does happen, but not always, as you've demonstrated).
Instead of partial integration, use the identity cos(2x) = 2cos^2(x)  1 <=> cos^2(x) = cos(2x)/2 + 1/2. 



#3
Dec2804, 11:22 AM

P: 104

Yes, I tried partial integration. And thank you for the help!
Cheers 



#4
Dec2804, 11:34 AM

Sci Advisor
HW Helper
P: 9,428

Integration of (Cos X)^2
integration by parts does work of course but only if in the second step, you refrain from undoing the work of the first step. this can be confusing and is actually easier to do by guesswork.
i.e. the derivative of sincos is cos^2  sin^2. but sin^2 + cos^2 = 1 is also easy to get as a derivative, namely it is the derivative of x. so the derivative of x + sincos is 2cos^2. now you are done after dividing by 2. of course you notice here that cos^2  sin^2 = cos(2x) is also coming in as in the trick suggested above, but here you do not have to know that trick. 



#5
Dec2804, 12:18 PM

Sci Advisor
HW Helper
P: 11,863

Check post number 7 (mine ) from this thread:sine&cosine squared
I think it sould be pretty clear... U have both the primitives/antiderivatives and the definite integrals of the 2 functions wrt to the limits \pi/2 and +\pi/2. Daniel. 



#6
Dec2804, 02:22 PM

P: 128

Make sure you read what Muzza said at the end. Just using the identity [tex]\text{cos}^2(x) = \frac{1}{2}(1+\text{cos}(2x))[/tex] makes this a very simple integral.




#7
Dec2804, 11:17 PM

Sci Advisor
HW Helper
P: 9,428

but the point is not everyone has this identity at their disposal.




#8
Dec2804, 11:22 PM

P: 646

Really!! I thought it was one of the basic identities in trigonometry usually referred to as the double angle formulae.
 AI 



#9
Mar1809, 06:16 PM

P: 21

hi,
i must integrate sin^2(x) by partial integration. i I've done this by taking the 2nd partial ingeral and substituting it with the original integral of sin^2(x); and calculate 0 =0. what is it that i'm doing wrong? what does mathwonk mean by : "integration by parts does work of course but only if in the second step, you refrain from undoing the work of the first step" ? 



#10
Mar1809, 08:11 PM

P: 74

why must you use integration by parts? When it's very easy to integrate using the power reducing formula.




#11
Mar1909, 03:11 AM

P: 21





#12
Mar1909, 03:39 AM

P: 21

got it : http://www.nevada.edu/~cwebster/Teac.../intparts.html
i guess mathwork meant that one shouldn't substitute the original integral into the second partial integral...? 



#13
Mar1909, 07:00 AM

P: 21

Doh!.. that's exactly the same method i was using, i still get 0=0 thus :(




#14
Mar1909, 09:22 AM

P: 2,159

So, it takes more than 4 years for Physics Forums to compute this integral?




#15
Mar1909, 09:28 AM

P: 21

i'm wondering how seriously i should take the "integrate by parts" bit, if i do a substitution in the second partial integral with a trigonometric identity, would that be considered cheating? 



#16
Mar1909, 04:27 PM

P: 74

you can integrate by parts as long as you use the Pythagorean identity. I don't see how that would be cheating.




#17
Mar1909, 04:37 PM

P: 21





#18
Mar1909, 04:45 PM

P: 74

Yes, never mind my earlier post I just made the process longer. You can absolutely do it by parts alone. Scratch that I messed up...Give me a minute



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