請教定積分法3題

∫(t^3 - 8) / (t - 2) dt
= ∫(t - 2)(t^2 + 2t + 4) / (t - 2) dt
= ∫(t^2 + 2t + 4) dt
= [(1/3)t^3 + t^2 + 4t]
= [(1/3)(1)^3 + (1)^2 + 4(1)] - [(1/3)(0)^3 + (0)^2 + 4(0)]
= 16/3

∫(1 + cosx)/(1 - cos2x) dx
= ∫(1 + cosx)/(2sin^2 x) dx
= ∫[1/(2sin^2 x) + cosx/(2sin^2 x)] dx
= ∫[(1/2)cosec^2 x + (1/2)cosecx cotx] dx
= -(1/2)[cotx + cosecx]
= -(1/2){[cot(pi/2) + cosec(pi/2)] - [cot(pi/3) + cosec(pi/3)]}
= [sqrt(3) - 1]/2

∫[(cos2x)^2 / (1 + cos2x) + (sin2x)^2 / (1 + cos2x)] dx
= ∫[(cos2x)^2 + (sin2x)^2] / (1 + cos2x) dx
= ∫1 / (1 + cos2x) dx
= ∫1 / (2cos^2 x) dx
= (1/2) ∫sec^2 x dx
= (1/2) [tan x]
= (1/2) [tan(pi/4) - tan(0)]
= 1/2