Evaluate indefinite integral $\int e^{-3t} \cos(2-\sqrt{3}t)dt $ using complex exponential

Question

$$\int e^{-3t} \cos(2-\sqrt{3}t)dt $$ I got to here $$\int e^{-3t}\left(\frac{1}{2}\left(e^{2i-i\sqrt{3}t} + e^{2i+i\sqrt{3}t}\right)\right)dt$$ But I'm not sure how to go on from here

Answer 1

Hint: for real $a,b$ with $(a,b) \ne (0,0)$ we have

$\int e^{at}\cos(bt) dt= \int Re(e^{(a+ib)t}) dt= Re(\int e^{(a+ib)t} dt=Re(\frac{1}{a+ib}e^{(a+ib)t}) $.

Answer 2

What is the problem? $$\int e^{-3t}\left(\frac{1}{2}\left(e^{2i-i\sqrt{3}t} + e^{2i+i\sqrt{3}t}\right)\right){\rm d}t=\int \left(\frac{1}{2}\left(e^{-3t}e^{2i-i\sqrt{3}t} + e^{-3t}e^{2i+i\sqrt{3}t}\right)\right){\rm d}t=\int\frac{1}{2}\left(e^{-3t+2i-i\sqrt{3}t} + e^{-3t+2i+i\sqrt{3}t}\right){\rm d}t=\frac{1}{2}\int e^{-3t+2i-i\sqrt{3}t}{\rm d}t + \frac{1}{2}\int e^{-3t+2i+i\sqrt{3}t}{\rm d}t=\frac{1}{2}\frac{e^{-3t+2i-i\sqrt{3}t}}{-3t+2i-i\sqrt{3}t}+\frac{1}{2}\frac{e^{-3t+2i+i\sqrt{3}t}}{-3t+2i+i\sqrt{3}t}$$ The denominators in the last expression have equal absolute values. If you expand both fractions by complex adjoint of their denominator you will be able to rewrite the result in terms of goniometric functions.