Using Complex exponential/Euler's Method to integrate, bizarre case

Question

$ \int e^{-3t} cos(2-\sqrt 3 t) dt $

I have been asked to evaluate that using complex exponential/euler's method. I have done many similar questions but all of them had something like (cos3x), sin(5t) etc. This is the first time I have come across a question where its of the format (cos a+bx) and cannot understand how to deal with the extra term. I have been stuck for a while now and have thus decided to ask the community for help, please help me out.

Answer 1

It's just like a complex constant thats all

$$\int e^{-3t} cos(2-\sqrt 3 t) dt $$

$$\int \frac {e^{-3t}} 2 (e^{+i(2-\sqrt 3 t)} + e^{-i(2-\sqrt 3 t))})dt $$

$$\int \frac {e^{-3t}} 2 e^{+i(2-\sqrt 3 t)} dt + \int \frac {e^{-3t}} 2 e^{-i(2-\sqrt 3 t)} dt $$

$$ e^ {2i} \int \frac {e^{-3t}} 2 e^{- i\sqrt 3 t} dt + e^ {-2i} \int \frac {e^{-3t}} 2 e^{i\sqrt 3 t} dt $$

$$ \frac {e^ {2i}} 2 \int {e^{-t ( 3 + i\sqrt 3)}} dt + \frac {e^ {-2i}} 2 \int {e^{-t( 3- i\sqrt 3 )}} dt )$$ Then integrate ...

Answer 2

Consider the more general case$$I=\int e^{at} \cos(b+ct) dt$$ and start with

$$b+ct =x\implies t=\frac{x-b}c\implies dt=\frac{dx}c$$ which makes $$I=\frac {e^{-\frac{ab}c}}c\int e^{\frac{a}c x}\cos(x) \,dx$$ At this point, you could consider $$J=\frac {e^{-\frac{ab}c}}c\int e^{\frac{a}c x}\sin(x) \,dx$$ which makes $$I+iJ=\frac {e^{-\frac{ab}c}}c\int e^{(\frac{a}c+i) x}\,dx=\frac {e^{-\frac{ab}c}}c \frac{c }{a+i c}e^{\frac{x (a+i c)}{c}}$$ $$I-iJ=\frac {e^{-\frac{ab}c}}c\int e^{(\frac{a}c-i) x}\,dx=\frac {e^{-\frac{ab}c}}c \frac{c }{a-i c}e^{\frac{x (a-i c)}{c}}$$ and combine them to get after simplifications $$I=\frac{ a \cos (x)+c \sin (x)}{a^2+c^2}\,e^{\frac{a (x-b)}{c}}$$ $$J=\frac{ a \sin (x)-c \cos (x)}{a^2+c^2}\,e^{\frac{a (x-b)}{c}}$$ Back to $t$, this would give $$I=\int e^{at} \cos(b+ct)\, dt=\frac{ a \cos (b+c t)+c \sin (b+c t)}{a^2+c^2}\,e^{a t}$$ $$J=\int e^{at} \sin(b+ct)\,dt=\frac{ a \sin (b+c t)-c \cos (b+c t)}{a^2+c^2}\,e^{a t}$$

Edit

We could even make it more complex and using exactly the same procedure get $$I=\int e^{at+d} \cos(b+ct)\, dt=\frac{ a \cos (b+c t)+c \sin (b+c t)}{a^2+c^2}\,e^{a t+d}$$ $$J=\int e^{at+d} \sin(b+ct)\, dt=\frac{ a \sin (b+c t)-c \cos (b+c t)}{a^2+c^2}\,e^{a t+d}$$