求不定积分∫e^xsin^2xdx

2个回答

  • ∫ e^xsin²x dx

    =(1/2)∫ e^x(1-cos2x) dx

    =(1/2)e^x - (1/2)∫ e^xcos2x dx (1)

    下面计算:

    ∫ e^xcos2x dx

    =∫ cos2x d(e^x)

    分部积分

    =e^xcos2x + 2∫ e^xsin2x dx

    =e^xcos2x + 2∫ sin2x d(e^x)

    再分部积分

    =e^xcos2x + 2e^xsin2x - 4∫ e^xcos2x dx

    将 -4∫ e^xcos2x dx 移项与左边合并后除以系数

    得:∫ e^xcos2x dx = (1/5)e^xcos2x + (2/5)e^xsin2x + C

    将上式代入(1)得

    ∫ e^xsin²x dx = (1/2)e^x - (1/10)e^xcos2x - (1/5)e^xsin2x + C

    若有不懂请追问,如果解决问题请点下面的“选为满意答案”.