Question

两道微积分题目，求解，万分感谢！

Answer 1

1）dA/dt=∫e^(-3t)cos2tdt
=1/2*∫e^(-3t)d(sin2t)
=1/2*[e^(-3t)sin2t+3*∫e^(-3t)sin2tdt]
=1/2*e^(-3t)sin2t-3/4*∫e^(-3t)d(cos2t)
=1/2*e^(-3t)sin2t-3/4*[e^(-3t)cos2t+3∫e^(-3t)cos2tdt]
=1/2*e^(-3t)sin2t-3/4*e^(-3t)cos2t-9/4*∫e^(-3t)cos2tdt
dA/dt=∫e^(-3t)cos2tdt
=2/13*e^(-3t)sin2t-3/13*e^(-3t)cos2t+C 其中C是任意常数
A=2/13*∫e^(-3t)sin2tdt-3/13*∫e^(-3t)cos2tdt+Ct
=2/13*[2/3*∫e^(-3t)cos2tdt-1/3*e^(-3t)sin2t]-3/13*∫e^(-3t)cos2tdt+Ct
=-5/39*[2/13*e^(-3t)sin2t-3/13*e^(-3t)cos2t+B]-2/39*e^(-3t)sin2t+Ct
=5/169*e^(-3t)cos2t-12/169*e^(-3t)sin2t+Ct+B
=1/169*e^(-3t)*(5cos2t-12sin2t)+Ct+B 其中B、C都是任意常数

2）令x=sect dx=tantsectdt
原式=∫tantsect/tan^5tdt
=∫sect/tan^4tdt
=∫cos^3t/sin^4tdt
=∫ctg^3tcsctdt
=-∫ctg^2td(csct)
=-ctg^2tcsct+∫csctd(ctg^2t)
=-ctg^2tcsct-2∫ctgtcsc^3tdt
=-ctg^2tcsct+2∫csc^2td(csct)
=2/3*csc^3t-ctg^2tcsct+C
=csct*(2/3*csc^2t-ctg^2t)+C
=x/√(x^2-1)*[2x^2/(3x^2-3)-1/(x^2-1)]+C
=x/√(x^2-1)*(2x^2-3)/(3x^2-3)+C 其中C是任意常数

好难打。希望能帮到你