求下列1/5-4sinx+3cosx的不定积分
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半角代换 tan(x/2) = u,
sinx = 2u/(1+u^2), cosx= (1-u^2)/(1+u^2), dx = 2du/(1+u^2)
∫[1/(5-4sinx+3cosx)]dx = ∫2udu/[5(1+u^2)-8u+3(1-u^2)]
= ∫udu/(4-4u+u^2) = ∫udu/(u-2)^2 = ∫[1/(u-2)+2/(u-2)^2]du
= ln|u-2|- 2/(u-2) + C = ln|tan(x/2)-2|- 2/[tan(x/2)-2] + C
sinx = 2u/(1+u^2), cosx= (1-u^2)/(1+u^2), dx = 2du/(1+u^2)
∫[1/(5-4sinx+3cosx)]dx = ∫2udu/[5(1+u^2)-8u+3(1-u^2)]
= ∫udu/(4-4u+u^2) = ∫udu/(u-2)^2 = ∫[1/(u-2)+2/(u-2)^2]du
= ln|u-2|- 2/(u-2) + C = ln|tan(x/2)-2|- 2/[tan(x/2)-2] + C
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