Fourier Transform Problem - absolute value, time-saving tricks, etc I am given the following signal:

$$[e^{-at}cos(w_{o}t)]u(t),\ a&gt;0$$

Then I am told to find the Fourier Transform, which tells me I need an answer of the form: 
$$X(jw)=\int_{-\infty}^\infty \! x(t)e^{-jwt} \, \mathrm{d}t.$$

I know I can reset the bounds of my integral with the unit step function, so my equation becomes
$$X(jw)=\int_{0}^\infty \! e^{-at}cos(w_ot)e^{-jwt} \, \mathrm{d}t.$$

From here, can I essentially solve this out and get a correct answer, keeping $w$ and $w_o$ as separate variables?

I also know I can solve it by using the relation 
$$x(t)=a(t)b(t)\xrightarrow{\mathscr{F}} X(jw)=\frac{1}{2\pi}A(jw)*B(jw)$$

So, essentially, I can figure out a transform for each part and convolve to find my answer?

From the book examples, it seems $cos(w_ot)$ can be broken down into $\pi\delta(w-w_o)+\pi\delta(w+w_o).$ If we convolve this with the result from the transform of $e^{-at}u(t)$, a correct answer should be obtained. 

It is known that $X(jw)$ when $x(t)=e^{-at}$ is $\frac{1}{(a+jw)}$

Therefore, since impulses sift through the other function in a convolution to get the nonzero values, is the following a correct result? $$\frac{1}{2\pi}(\pi)(\frac{1}{(a+j(w-w_o))}+\frac{1}{(a+j(w+w_o))})$$

$$=\frac{1}{2}(\frac{1}{(a+j(w-w_o))}+\frac{1}{(a+j(w+w_o))})$$

Thank you, sorry for the long question!
As you suggested, you could simply solve the integral using $\cos(\omega_0t)=(e^{j\omega_0t}+e^{-j\omega_0t})/2$. But as you've also noted, with this type of problems there is usually some smart way using known transform pairs. And what you suggested appears to me a very sane approach: convolve the known transform of $\cos(\omega_0t)$ with the known transform of $e^{-at}u(t)$ and you're done. I would recommend to you to cross-check your result by solving the integral, which in this case is also quite straight-forward. But I can assure you that your result looks good.