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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 11:47:38 IST
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make given integral as
{(3+2cosx)/3sinx}d(1/2+3cosx)
apply parts
the integral trem comes out as
integarl[ {(2sin^2x+(3+2cosx)cosx)/sin^2x}*dx/(3+2cosx)]
= (3+2cosx/sin^2x)*dx/3+2cosx= cosec^2xdx
which is direct
ofcourse there will b some 1/3 or something outside as 1/3*integral..but thts all not important here
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Nitwit Blubber Odment Tweak
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this reply: 10 points
(with 2 
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