Integral of sin(theta)/(cos^2(theta)-2cos(theta))d(theta)

Answer 1

∫sinς/(cos²ς - 2cosς)dξ; let do substitution u = cosς, du = - sinςdς; and we have : ∫sinς/(cos²ς - 2cosς)dς =

-∫du/(u² 2u) = - ∫du/[(u - 1)² - 1} = -∫1/2{[1/(u-1) - 1] - 1/[(u - 1) + 1]}d(u - 1) = 1/2lnIu/(u - 2)I + C

= 1/2lnIcosς/(cosς -2)I + C