求sinx的三次方除以(sinx+cosx)的不定积分
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I = ∫(sinx)^3 dx/(sinx+cosx) = (1/√2)∫(sinx)^3 dx/sin(x+π/4), 记 u = x+π/4，
I = (1/√2)∫[sin(u-π/4)]^3 du/sinu = (1/4)∫(sinu-cosu)^3 du/sinu
= (1/4)∫[(sinu)^2-3sinucosu+3(cosu)^2-(cosu)^3/sinu] du
= (1/4){∫[2 + cos2u -3sinucosu]du - ∫[1-(sinu)^2]/sinu] dsinu}
= (1/4)[2u + (1/2)sin2u -(3/2)(sinu)^2] - ln|siu| + (1/2)(sinu)^2] + C1
= (1/4)[2u + (1/2)sin2u - (sinu)^2] - ln|sinu| ] + C1
= (1/4)[2x+π/2 + (1/2)cos2x - (sin(x+π/4))^2] - ln|sin(x+π/4)| ] + C1
= (1/4)[2x + (1/2)cos2x - (sin(x+π/4))^2] - ln|sin(x+π/4)| ] + C
I = (1/√2)∫[sin(u-π/4)]^3 du/sinu = (1/4)∫(sinu-cosu)^3 du/sinu
= (1/4)∫[(sinu)^2-3sinucosu+3(cosu)^2-(cosu)^3/sinu] du
= (1/4){∫[2 + cos2u -3sinucosu]du - ∫[1-(sinu)^2]/sinu] dsinu}
= (1/4)[2u + (1/2)sin2u -(3/2)(sinu)^2] - ln|siu| + (1/2)(sinu)^2] + C1
= (1/4)[2u + (1/2)sin2u - (sinu)^2] - ln|sinu| ] + C1
= (1/4)[2x+π/2 + (1/2)cos2x - (sin(x+π/4))^2] - ln|sin(x+π/4)| ] + C1
= (1/4)[2x + (1/2)cos2x - (sin(x+π/4))^2] - ln|sin(x+π/4)| ] + C
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