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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Jun 2007 22:26:56 IST
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  tan x
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ladder of success is never crowded at top |
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29 Jun 2007 22:37:53 IST
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log (1+ root 2 . sqaure root of tan x + tan x)
Please rate me if I am correct !
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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this reply: 5 points
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1 Jul 2007 19:14:24 IST
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well..... put tanx= t^2 => sec^2x dx = 2tdt => (1+ tan^2 x) dx= 2t dt => (1+t^4) dx = 2t dt => dx = 2t/(1+t^4) dt .: [ ] (tan x)1/2 dx = [ ] 2t^2/(1+t^4) dt now integrate using standard formulae.... Goutham Harsha (I find bliss in ignorance [^_^] )
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