∫e^3x*cos2xdx,∫coslnxdx,求不定积分,

Question

∫e^3x*cos2xdx,∫coslnxdx,求不定积分,

Answer 1

∫ e^(3x)cos(2x) dx
= (1/2)∫ e^(3x) dsin(2x)
= (1/2)e^(3x)sin(2x) - (1/2)(3)∫ e^(3x)sin(2x) dx
= (1/2)e^(3x)sin(2x) - (3/2)(- 1/2)∫ e^(3x) dcos(2x)
= (1/2)e^(3x)sin(2x) + (3/4)e^(3x)cos(2x) - (3/4)(3)∫ e^(3x)cos(2x) dx
(1 + 9/4)∫ e^(3x)cos(2x) dx = (1/4)e^(3x)(2sin2x + 3cos2x)
∫ e^(3x)cos(2x) dx = (1/13)[2sin(2x) + 3cos(2x)]e^(3x) + C
∫ cos(lnx) dx = ∫ xcos(lnx) · 1/x dx
= ∫ xcos(lnx) d(lnx) = ∫ x dsin(lnx)
= xsin(lnx) - ∫ sin(lnx) dx
= xsin(lnx) - ∫ xsin(lnx) d(lnx)
= xsin(lnx) + ∫ x dcos(lnx)
= xsin(lnx) + xcos(lnx) - ∫ cos(lnx) dx
2∫ cos(lnx) dx = x[sin(lnx) + cos(lnx)]
∫ cos(lnx) dx = (x/2)[sin(lnx) + cos(lnx)] + C