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

mathreadler
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randb
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  • Use that, for every complex number $z$, $$(e^{zt})'=ze^{zt}$$ hence, for every nonzero complex number $z$, $$\int e^{zt}dt=\frac1ze^{zt}+C$$ plus, of course, the trick that, for every real number $u$, $$\cos(u)=\Re(e^{iu})$$ – Did Aug 31 '17 at 09:35

2 Answers2

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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}) $.

Fred
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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.

Puzzled student
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