Question

∫e^axCosbxdx

求不定积分，可以1/b∫e^axd(Sinbx)这样吗？

Answer 1

正是这样做。
N = ∫ e^(ax)cos(bx) dx
= (1/b)∫ e^(ax) dsin(bx)
= (1/b)e^(ax)sin(bx) - (1/b)∫ sin(bx)[ae^(ax)] dx
= (1/b)e^(ax)sin(bx) - (a/b)(- 1/b)∫ e^(ax) dcos(bx)
= (1/b)e^(ax)sin(bx) + (a/b²)e^(ax)cos(bx) - (a/b²)∫ cos(bx)[ae^(ax)] dx
= (1/b²)[bsin(bx) + acos(bx)]e^(ax) - (a²/b²)N
(1 + a²/b²)N = [(a² + b²)/b²]N = (1/b²)[bsin(bx) + acos(bx)]e^(ax)
N = e^(ax)[bsin(bx) + acos(bx)]/(a² + b²)

同样地，先积e^(ax)也行
N = ∫ e^(ax)cos(bx) dx = (1/a)∫ cos(bx) de^(ax)
= (1/a)e^(ax)cos(bx) - (1/a)∫ e^(ax)[- bsin(bx)] dx
= ...

Answer 2

可以使用拼凑法，答案如图所示