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∫sin

解: 令lnx=t,则x=e^t ∫sin(lnx)dx =∫sintd(e^t) =e^t·sint-∫e^td(sint) =e^t·sint -∫costd(e^t) =e^t·sint -e^t·cost +∫e^td(cost) =(sint-cost)·e^t-∫sintd(e^t) 2∫sintd(e^t)=(sint-cost)·e^t ∫sintd(e^t)=½[sin(lnx)-cos(lnx)]·x+C ∫...

(1/4)(2x-sin2x)+C 解析: f(x) =sin²x =(1-cos2x)/2 ∫f(x)dx =∫[(1-cos2x)/2]dx =(1/2)∫(1-cos2x)dx =(1/2)[x-(1/2)sin2x]+C =(1/4)(2x-sin2x)+C

∫sin(lnx)dx =xsin(lnx)-∫xdsin(lnx) =xsin(lnx)-∫x*cos(lnx)*1/xdx =xsin(lnx)-∫cos(lnx)dx =xsin(lnx)-xcos(lnx)+∫xdcos(lnx) =xsin(lnx)-xcos(lnx)-∫x*sin(lnx)*1/xdx =xsin(lnx)-xcos(lnx)-∫sin(lnx)dx 所以2∫sin(lnx)dx=xsin(lnx)-xcos(lnx...

解: ∫sin²tdt =∫[1-cos(2t)]/2 dt =∫(1/2)dt -(1/2)∫cos(2t)dt =∫(1/2)dt -(1/4)∫cos(2t)d(2t) =t/2 -(1/4)sin(2t)+C

∫sin2xdx =(1/2)∫sin2xd(2x) =-(1/2)cos2x + C

∫sin(ax+b)dx =1/a∫sin(ax+b)d(ax+b) =1/a (-cos(ax+b))+c =-1/a cos(ax+b)+c

∫sin2xdx =1/2∫sin2xd2x =-1/2*cos2x+C

∫(sinx)^4 dx =(1/4)∫(1-cos2x)^2 dx =(1/4)∫[1-2cos2x+ (cos2x)^2] dx =(1/8)∫[3-4cos2x+ cos4x] dx =(1/8)[ 3x-3sin2x+(1/4)sin4x] + C

2sin²(ωt +ψ)=1-cos(2ωt +ψ) ∫sin²(ωt +ψ)dt =1/2t-1/4ωsin2(ωt+ψ)+C

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