Posted by anonymous on Sunday, January 6, 2008 at 6:09pm.
Well, I happen to know that the derivate of the tangent is sec^2 :)
(cosx)/(sin^3x) dx
Try f(x) = 1/sin^2 x
f' = - 2 sin x cos x /sin^4 x
see what happens?
let u = 2 x
then du = 2 dx
and we have
csc u cot u du/2 from u = pi/2 to pi/6
(1/sin u)(cos u/sin u) du/2
= (1/2) (cos u/sin^2 u) du
try 1/sin u
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