what is the integratin of 1/sin2x

Use the formula ∫cosec(x) dx = -cosec(x)cot(x), and let 2x=y

∫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)]

If you like the answer, please click YES below