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![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Apr 2008 16:09:40 IST
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/(1-tanx))
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25 Apr 2008 16:15:23 IST
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I think it is :-
^2/(1-tan^2x)) dx
dx)
%2B1/2(ln|sec%202x|)%20%2B%20c)
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25 Apr 2008 16:15:54 IST
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is it RITE?????????????
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25 Apr 2008 16:16:51 IST
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is the answer -log(cosx-sinx)??
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25 Apr 2008 16:18:24 IST
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show the working
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25 Apr 2008 16:24:52 IST
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expanding 1+tanx/1-tanx
we get
cosx+sinx/cosx-sinx
d/dx of cosx -sinx is -(sinx+cosx)
so let
cosx-sinx=t
then we get
 -dt/t
= -logt
=-log(cosx-sinx)
correct if i am wrong!!!
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25 Apr 2008 16:24:53 IST
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well
convert tan into sin and cos
u ll have
[cos x + sinx] / [cos x - sin x]
put denom as t
differentiate it
substitute
and get d answr
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I Tried So Hard And Got So Far
But In The End
It Doesnt Even Matter... |
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25 Apr 2008 16:28:29 IST
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thts rite but what about the 2nd post??? is tht rite??
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25 Apr 2008 16:52:14 IST
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u can wright it as tan(x+pi/4), then integrate it, u get log sec x+pi/4
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