| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Feb 2007 11:59:32 IST
|
|
|
Integrate w.r.t x  1/ ( sinx + xcosx) dx
|
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Feb 2007 12:22:28 IST
|
|
|
hey some one plz try out this ques. its a gud one..............
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Feb 2007 18:06:12 IST
|
|
|
We know sinx=(eix-e-ix)/(2i) and cosx=(eix+e-ix)/2 So given integral is  2idx/((e ix-e -ix)+xi(e ix+e -ix) =  2ie xdx/(e 2ix(1+xi)+(xi-1)) Put eix=t ieixdx=dt So =2  dt/(t 2(logt+1)+(logt-1)) Can anyone help me from here.Also tell me whether this is the right approach or not!!!
|
ADARSH
NITK Surathkal
|
this reply: 10 points
(with 2 
in 2 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Feb 2007 18:12:16 IST
|
|
|
wat u did is rite kab
|
TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
     
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Feb 2007 18:12:41 IST
|
|
|
hey kab i think now we ve 2 apply by parts................but it will b very complicated
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Feb 2007 18:14:59 IST
|
|
|
kab heres salute for u
|
TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
     
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Feb 2007 18:18:57 IST
|
|
|
Thanks.May be there is another method.Experts plz help!!!
|
ADARSH
NITK Surathkal
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Feb 2007 18:20:07 IST
|
|
|
sumone plzzzz try this ques
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Feb 2007 18:21:53 IST
|
|
|
madhuri how do u know that the kab's soln is correct...............ques is nt solved yet................but its a gud try by kab
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Feb 2007 18:26:33 IST
|
|
|
that is a very good try isn't it??
|
TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
     
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Mar 2007 20:38:58 IST
|
|
|
substituting further from kab's soln log t = u 1/t = du/dt this gives integ of 2 e ^u du / e^u *( e^2u -1 ) + u (e^2u + 1 ) now can any one help how to go further
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Mar 2007 21:22:42 IST
|
|
|
Dear sabya_sachi please check and conform whether it is a definite integral or indefinite. if definite please give the limits too.
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|