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∫sin^2(3X+1)dX
=-1/2∫[1-2sin^2(3X+1)-1]dX
=-1/2∫[cos(6X+2)-1]dX
=-1/2×1/6∫[cos(6X+2)-1]d(6X+2)
=-1/12[sin(6X+2)-(6X+2)]+C
=-1/12sin(6X+2)+1/2X+1/6+C
=-1/12sin(6X+2)+1/2X+C(C为常数)
=-1/2∫[1-2sin^2(3X+1)-1]dX
=-1/2∫[cos(6X+2)-1]dX
=-1/2×1/6∫[cos(6X+2)-1]d(6X+2)
=-1/12[sin(6X+2)-(6X+2)]+C
=-1/12sin(6X+2)+1/2X+1/6+C
=-1/12sin(6X+2)+1/2X+C(C为常数)
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∫ [sin(3x+1)]^2 dx
=(1/2) ∫ [1-cos(6x+2)] dx
=(1/2)[ x -(1/6)sin(6x+2) ] + C
=(1/2) ∫ [1-cos(6x+2)] dx
=(1/2)[ x -(1/6)sin(6x+2) ] + C
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追问
请问一下这是第一类换元积分法吗
追答
利用三角公式
cos2u = 1-2(sinu)^2
(sinu)^2 = (1/2)(1- cos2u)
带入 u=3x+1
得出
[sin(3x+1)]^2 =(1/2)[1-cos(6x+2)]
∫ [sin(3x+1)]^2 dx
=(1/2)∫ [1-cos(6x+2)] dx
=(1/2)[ x -(1/6)sin(6x+2) ] + C
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