求∫e∧x sinx∧2 dx的解题过程,？
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∫ e^x * sin²x dx
= ∫ e^x * (1 - cos2x)/2 dx
= (1/2)e^x - (1/2)N
N = ∫ e^xcos2x dx = (1/2)∫ e^x d(sin2x)
= (1/2)e^xsin2x - (1/2)∫ e^xsin2x dx
= (1/2)e^xsin2x + (1/4)∫ e^x d(cos2x)
= (1/2)e^xsin2x + (1/4)e^xcos2x - (1/4)N
(5/4)N = (1/4)(2sin2x + cos2x)e^x
N = (1/5)(2sin2x + cos2x)e^x + C
原式 = (1/2)e^x - (1/5)e^xsin2x - (1/10)e^xcos2x + C,2,
= ∫ e^x * (1 - cos2x)/2 dx
= (1/2)e^x - (1/2)N
N = ∫ e^xcos2x dx = (1/2)∫ e^x d(sin2x)
= (1/2)e^xsin2x - (1/2)∫ e^xsin2x dx
= (1/2)e^xsin2x + (1/4)∫ e^x d(cos2x)
= (1/2)e^xsin2x + (1/4)e^xcos2x - (1/4)N
(5/4)N = (1/4)(2sin2x + cos2x)e^x
N = (1/5)(2sin2x + cos2x)e^x + C
原式 = (1/2)e^x - (1/5)e^xsin2x - (1/10)e^xcos2x + C,2,
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