What is the integratin of 1/sin2x?, Mathematics

∫1/sin2x dx = ∫cosec2x dx = 1/2 log[cosec2x - cot2x] + c = 1/2 log[tan x] + c

Detailed derivation of

∫cosec x dx = ∫cosec x(cosec x - cot x)/(cosec x - cot x) dx

= ∫(cosec2x - cosecxcotx)/(cosecx - cotx) dx

= log[cosecx - cotx] = log[(1-cosx)/sinx]

= log[2sin2(x/2)/2sin(x/2)cos(x/2)]

= log[tan(x/2)]

Posted Date: 3/11/2013 3:07:54 AM | Location : United States







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