Help!!!by beanryu Tags: None 

#1
Mar1506, 12:37 AM

P: 96

can anyone do the following integration?! PLEASE!!!
S=integral sine S(lnz)^2 



#2
Mar1506, 12:43 AM

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1. Yes, we can do them, but we are not going to do your hw for you, as that will not help you in any way. Please read the forum guidelines on hw help here:
http://www.physicsforums.com/showthread.php?t=94383 The gist of it is that you must post your work and show that you have made a decent attempt, but got stuck somewhere. Then we can help point you in the right direction. 2. Your question is not very clear, and your notation not very good. For the first integral, do you mean: [tex] \int{\sin x \ dx} [/tex] If so, are you serious??? You don't know this integral? As for this one: [tex] \int{(\ln z)^2 \ dz} [/tex] Did you attempt integration by parts? 



#3
Mar1506, 12:44 AM

P: 96

yes i dont no!!!!
i tried it for like the whole half day i dont have too much time please HELP!!!! 



#4
Mar1506, 12:46 AM

P: 96

Help!!!
(zlnzz)(z)(zlnzz)(1/z)????????




#5
Mar1506, 12:49 AM

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Sorry, I don't believe you. If you attempted it for half a day, why not post your work as I suggested? Also, what exactly did you "try" for the first one? I can't imagine you getting very far putting pencil to paper if you don't already know the answer from beforehand. The integrals and derivatives of sine and cosine functions are something that you just look at the derivation/proof for once and then remember the result after that. Everybody knows them; there is no computation involved. This integral is given in any basic calculus textbook and so it makes no sense whatsoever that this integral would be given for hw. You never answered my question for the second one. Did you attempt integration by parts? 



#6
Mar1506, 12:54 AM

P: 347

I think by "S=integral sine" they meant that S will represent the integral sign. So they're trying to find [itex]\int (\ln{z})^2 \, dz[/itex], and as you suggested by parts would be a good idea.




#7
Mar1506, 12:55 AM

P: 96

....................................................................... ......... i m dead




#8
Mar1506, 12:56 AM

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#9
Mar1506, 12:56 AM

P: 347

Notice that [itex] (\ln{z})^2 = (\ln{z})^2 * 1[/itex]. Can you see how integration by parts applies here?




#10
Mar1506, 12:57 AM

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#11
Mar1506, 01:00 AM

P: 96

thanx alot, but time out for me .... THANX ALOT THOUGH



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