计算：1.∫（x^2-1)sin2xdx 2.∫cos(lnx)dx 3.已知一曲线过点（1,2）且任一点的切线斜率为3

3.
点（1,2)
y'=3x^2
y=∫3x^2dx=x^3+C
x=1,y=2 ,C=1,
y=x^3+1
2
∫cos(lnx)dx
lnx=u x=e^u
=∫cosude^u=e^ucosu+∫e^usinudu=e^ucosu+∫sinude^u=e^ucosu+e^usinu-∫cosue^udu
=e^ucosu+e^usinu-∫cosude^u
2∫cosude^u=e^u(cosu+sinu)
∫cosue^u=(1/2)e^u(cosu+sinu)
=(1/2)x(coslnx+sin(lnx)
1
∫(x^2-1)sin2xdx
=∫x^2sin2xdx+(1/2)cos2x
=(1/2)∫x^2d(-cos2x)+(1/2)cos2x
=(-1/2)x^2cos2x+(1/2)∫cos2xdx^2+(1/2)cos2x
=(-1/2)x^2cos2x+∫xcos2xdx+(1/2)cos2x
=(-1/2)x^2cos2x+(1/2)∫xdsin2x+(1/2)cos2x
=(-1/2)x^2cos2x+(1/2)xsin2x-(1/2)∫sin2xdx+(1/2)cos2x
=(-1/2)x^2cos2x+(1/2)xsin2x+(3/4)cos2x+C

1.∫（x^2-1)sin2xdx=∫x^2sin2xdx-∫sin2xdx=-（x^2cos2x)/2+(xsin2x)/2+3cos2x/4+c
2.∫cos(lnx)dx=(1/2)x[sin(lnx)+cos(lnx)]+c
设lnx=t
x=e^t
dx=e^tdt
∫cos(lnx)dx=∫e^tcostdt=∫e^tdsint=e^tsint-...

1.∫（x^2-1)sin2xdx=∫x^2sin2xdx-∫sin2xdx=-（x^2cos2x)/2+(xsin2x)/2+3cos2x/4+c
2.∫cos(lnx)dx=(1/2)x[sin(lnx)+cos(lnx)]+c
设lnx=e^u
dx=e^udu
∫cos(lnx)dx=∫e^ucosudu=∫e^udsinu=e^usinu-∫e^usinu...