求解这道积分题
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以下答案来自数理数一数二团，希望可以帮到你！
这个好解着呢，分部积分法。
我只做不定积分，定积分这里不好输，自己做。
∫e^(-3t)cos2tdt
=(-1/3)∫cos2td[e^(-3t)]
=(-1/3)[e^(-3t)]cos2t-(-1/3)∫[e^(-3t)]dcos2t
=(-1/3)[e^(-3t)]cos2t-(2/3)∫sin2t[e^(-3t)]dt
=(-1/3)[e^(-3t)]cos2t+(2/9)∫sin2td[e^(-3t)]
=(-1/3)[e^(-3t)]cos2t+(2/9)[e^(-3t)]sin2t-(2/9)∫[e^(-3t)]dsin2t
=(-1/3)[e^(-3t)]cos2t+(2/9)[e^(-3t)]sin2t-(4/9)∫[e^(-3t)]cos2tdt
∴[1+(4/9)]∫[e^(-3t)]cos2tdt=(-1/3)[e^(-3t)]cos2t+(2/9)[e^(-3t)]sin2t
∴∫[e^(-3t)]cos2tdt=[e^(-3t)][(-3/13)cos2t+(2/13)sin2t]+C
这个好解着呢，分部积分法。
我只做不定积分，定积分这里不好输，自己做。
∫e^(-3t)cos2tdt
=(-1/3)∫cos2td[e^(-3t)]
=(-1/3)[e^(-3t)]cos2t-(-1/3)∫[e^(-3t)]dcos2t
=(-1/3)[e^(-3t)]cos2t-(2/3)∫sin2t[e^(-3t)]dt
=(-1/3)[e^(-3t)]cos2t+(2/9)∫sin2td[e^(-3t)]
=(-1/3)[e^(-3t)]cos2t+(2/9)[e^(-3t)]sin2t-(2/9)∫[e^(-3t)]dsin2t
=(-1/3)[e^(-3t)]cos2t+(2/9)[e^(-3t)]sin2t-(4/9)∫[e^(-3t)]cos2tdt
∴[1+(4/9)]∫[e^(-3t)]cos2tdt=(-1/3)[e^(-3t)]cos2t+(2/9)[e^(-3t)]sin2t
∴∫[e^(-3t)]cos2tdt=[e^(-3t)][(-3/13)cos2t+(2/13)sin2t]+C
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