Here is an integral that I want to see a different approach:
$$\int_0^{\infty} \frac{\sin 2t}{1+t^3}\, {\rm d}t$$
Well, for someone who is deeply aware of the exponential integral function and the cosine and sine integral functions well the answer would come pretty quickly. Because starting from the Laplace Transform of $\dfrac{1}{t+a}$ , that is:
$$\mathcal{L}\left \{ \frac{1}{t+a} \right \}= e^{as}{\rm Ei}\left ( as \right )$$
one can easily then deduce that:
$$\int_{0}^{\infty}\frac{e^{i\omega t}}{t+a}\, {\rm d}t= e^{-i a \omega}{\rm Ei}\left ( -i a\omega \right )$$
and the go one with his business using know identities that involve the exponential integral function, the cosine and sin integral function.
Then manipulating the denominator a bit (writing it as a finite sum etc) we will actually be in position to compute the more general integral:
$$\int_{0}^{\infty}\frac{\sin\omega t}{1+t^{n}}\,{\rm d}t, \;\; n \in \mathbb{N}$$
For someone however , who is not that deeply connected with these kind of functions, as me for example (i usually avoid using them in evaluations of integrals), these are completely inunderstandable (some part of them anyway). So, is there any other way to work around this integral? Residues or contour integration may actually work here. But the tricky part is to find the right contour or the right kernel. Unless there is no other way of going around it.