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How would one take the fourier cosine transform of $e^{-x}cos(x)$ ?

They've given that for a function f(x) = $e^{-bx}$ where b>0, the fourier cosine transform is given by $\frac{b}{k^2+b^2}$

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    Do you have a definition of the Fourier transform? show what you have tried? If they give you that hint then you should be able to resolve your function to something similar. – 2017-02-21
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    Yeah i have the definition.. I didn't reallly know where to start cos the only think i thought of was to compute the entire thing but that seemed to be the wrong way to go about it seeing as i wouldn't be using the hint then – 2017-02-21

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Letting $b=1-i$ we can see that

$$\begin{align} \int_0^\infty e^{-x}\cos(x)\cos(kx)\,dx&=\text{Re}\left(\int_0^\infty e^{-(1-i)x}\cos(kx)\,dx\right)\\\\ &=\text{Re}\left( \frac{1-i}{k^2+(1-i)^2}\right)\\\\ &=\frac{k^2+2}{k^4+4} \end{align}$$

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    (+1), but it should be stated that this is a "unilateral" Fourier cosine transform, since $e^{-x}\cos(x)$ is not a function in $L^1(\mathbb{R})$ but just in $L^1(\mathbb{R}^+)$. Maybe $e^{-\color{red}{|x|}}\cos(x)$ was really intended by the OP. – 2017-02-21
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    @jackd'aurizio Inasmuch as the OP give the CFT for $e^{-bx}$, and without absolute value on $x$, I assumed it was unilateral. Note that with an absolute value on $x$, the bilateral CFT would have been twice the result in the OP. And thank you for the (+1)! – 2017-02-21