∫e^(ax)cos(bx+c)dx=?

I = ∫exp(a*x)*cos(b*x + c)*dx
u = exp(a*x) => du = a*exp(a*x)*dx
dv = cos(b*x + c)*dx => v = (1/b)*sin(b*x + c)

I = exp(a*x)*(1/b)*sin(b*x + c) - (a/b)*∫exp(a*x)*sin(b*x + c)*dx
u = exp(a*x) => du = a*exp(a*x)*dx
dv = sin(b*x + c)*dx => v = - (1/b)*cos(b*x + c)

I = exp(a*x)*(1/b)*sin(b*x + c)
- (a/b)*[- exp(a*x)*(1/b)*cos(b*x + c) + (a/b)*I]
=> I*[1 + (a/b)^2] = exp(a*x)*(1/b)*[sin(b*x + c) + (a/b^2)*cos(b*x + c]
=> I = exp(a*x)*[b*sin(b*x + c) + a*cos(b*x + c)]/(a^2 + b^2)

噗...錯很大...懶的改了...直接改到意見好了- -

部分積分公式:∫udv = uv - ∫vdu

設u = e^(ax) => du = a * e^(ax)
dv = cos(bx+c) => v = (1/b)sin(bx+c)


=>∫e^(ax)cos(bx+c)dx
=[e^(ax)]*[(1/b)sin(bx+c)] -∫[(1/b)sin(bx+c)]*[a * e^(ax)] dx
=(1/b)*e^(ax)*sin(bx+c) -(a/b) ∫e^(ax)sin(bx+c) dx <--再做一次部分積分

2011-12-06 20:22:52 補充：
設u = e^(ax) => du = a * e^(ax)
dv = sin(bx+c) => v = (-1/b)cos(bx+c)

=>(1/b)*e^(ax)*sin(bx+c) -(a/b) ∫e^(ax)sin(bx+c) dx

= (1/b)*e^(ax)*sin(bx+c)
-(a/b) { [e^(ax)]*[(-1/b)cos(bx+c)] -∫[(-1/b)cos(bx+c)]*[a * e^(ax)] dx }

2011-12-06 20:28:33 補充：
=(1/b)*e^(ax)*sin(bx+c) + [a*e^(ax)*cos(bx+c)] / (b^2)
-[(a^2) / (b^2)] * ∫e^(ax)cos(bx+c) dx -----------1式


1式=原式∫e^(ax)cos(bx+c)dx

=>{(1+[a^2 / b^2])}∫e^(ax)cos(bx+c)dx
=(1/b)*e^(ax)*sin(bx+c) + [a*e^(ax)*cos(bx+c)] / (b^2)

2011-12-06 20:36:39 補充：
=>∫e^(ax)cos(bx+c)dx
={(1/b)*e^(ax)*sin(bx+c) + [a*e^(ax)*cos(bx+c)] / (b^2) } / [1+(a^2 / b^2)]

=[b(e^ax)sin(bx+c) + a(e^ax)cos(bx+c)] / (a^2+b^2)

回答(1)答案也許有誤，請參考:
http://www.wolframalpha.com/input/?i=\\[Integral]\\[ExponentialE]^%28a+x%29*Cos[b+x%2Bc+]\\[DifferentialD]x

Of course not!