Hint: Consider the more general integral
$$ \int_{0}^{\infty} x\sin(x) e^{-sx} dx $$
which is the Laplace transform of $x\sin(x)$. You need to use the following property
$$ L(xf(x))= -F'(s). $$
So, basically, you need to find only the Laplace transform of $\sin(x)$
Suppose $t>0$. (If $t \leq 0$, we can keep pushing the contour to the right and make the integral as small as we like, so the inverse transform gives $0$ there as it should). We can then change variables immediately to get the $t$-dependence out of the way: if $z=st$, the integral becomes
$$ t^p \frac{\Gamma(p+1)}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{e^{z}}{z^{p+1}} \, dz. $$
It remains to evaluate the constant $ \frac{\Gamma(p+1)}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{e^{z}}{z^{p+1}} \, dz$. We sweep the contour around to wrap around the branch point at $s=0$ (the integrals over the quarter-circles vanish in the limit by Jordan's lemma). The integral becomes
$$ \frac{\Gamma(p+1)}{2\pi i} \int_{\gamma} \frac{e^{z}}{z^{p+1}} \, dz, $$
where $\gamma$ is composed of a line just below the imaginary axis starting at $-\infty$ and ending at $-\varepsilon$, a circle of radius $\varepsilon$ around the branch point at $z=0$, and the line from $-\varepsilon$ back to $-\infty$ just above the imaginary axis. We could try and calculate this directly, but we'll be better off integrating by parts a few times to make the singularity at the branch point better behaved, so we can just take the limits directly. Also, integration by parts is easy with this new contour, because the exponential decays and causes the endpoint terms to vanish. We have
$$ \frac{\Gamma(p+1)}{2\pi i} \int_{\gamma} z^{-p-1}e^{z} \, dz = - \frac{\Gamma(p+1)}{2\pi i \,p} \int_{\gamma} -z^{1-p-1}e^{z} \, dz = \frac{\Gamma(p+1-1)}{2\pi i} \int_{\gamma} z^{1-p-1}e^z \, dz \\
= \dotsb = \frac{\Gamma(p+1-n)}{2\pi i} \int_{\gamma} z^{n-p-1} e^{z} \, dz $$
If $n-p>0$, the integrand no longer diverges at $z=0$ (but still has a branch point), so the integral over the circle vanishes as $\varepsilon \to 0$.
Therefore,
$$ \int_{\gamma} z^{n-p-1} e^{z} \, dz = \int_{-\infty}^0 z^{n-p-1} e^{z} \, dz + \int_0^{-\infty} z^{n-p-1} e^{z} \, dz $$
Changing variables to $z=xe^{-i\pi}$ in the first and $z=xe^{i\pi}$ in the second gives
$$ \int_{-\infty}^0 z^{n-p-1} e^{z} \, dz + \int_0^{-\infty} z^{n-p-1} e^{z} \, dz = -\int_0^{\infty} e^{-i\pi(n-p)} x^{n-p-1} e^{-x} \, dx + \int_0^{\infty} e^{i\pi(n-p)} x^{n-p-1} e^{-x} \, dx \\
= 2i\sin{(n-p)\pi} \int_0^{\infty} x^{n-p-1} e^{-x} \, dx = 2i \Gamma(n-p)\sin{(n-p)\pi}. $$
Thus the whole lot evaluates to
$$ t^p \frac{\Gamma(p+1-n)}{2\pi i} 2i \Gamma(n-p)\sin{(n-p)\pi} = \frac{\sin{(n-p)\pi}}{\pi} \Gamma(n-p)\Gamma(p+1-n) $$
But we have the reflection formula
$$ \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{\pi z}}, $$
so the constant $ \frac{\sin{(n-p)\pi}}{\pi} \Gamma(n-p)\Gamma(p+1-n) $ is actually $1$. Therefore the inverse Laplace transform gives $t^p$, as it should.
Best Answer
To actually evaluate the inverse Laplace transform integral, we must perform an integration in the complex plane. Many times, this means using the theorems of complex analysis, notably Cauchy's residue theorem. Because many students are introduced to the Laplace transform before they take complex analysis - many times, such students won't ever take a course in complex analysis - most first time lessons focus on using tables, to your chagrin.
If you are comfortable with integration in the complex plane, however, evaluating the ILT provides a terrific exercise for your contour integration. For example, consider a LT $\hat{f}(s)$ having poles in the complex plane. Rather then try to actually evaluate the ILT integral directly, we consider the integral over the closed contour $C$:
$$\frac{1}{i 2 \pi} \oint_C ds \, \hat{f}(s) \, e^{s t}$$
What is $C$? It turns out that our choice of $C$ will depend on the sign of $t$. When $t>0$, we will need to add a circular contour to the left of the vertical line of integration of the ILT. When $t<0$, we close to the right instead. The reason for this is that we need the integral over the circular contour to vanish as the radius $T$ of the circle approaches $\infty$. To see this, parametrize the integral there as $s = T e^{i \phi}$; then the integral over a leftward-circular contour looks like
$$i T\int_{\pi/2}^{3 \pi/2} d\phi \, e^{i \phi} \, \hat{f}(T e^{i \phi}) e^{t T \cos{\phi}} e^{i t T \sin{\phi}}$$
Note that $\cos{\phi} < 0$ within the integration region; thus the integral only vanishes when a) $t > 0$, and b) $\lim_{T \to \infty} T \hat{f}(T) = 0$.
This brings us to my next point: causality. The ILT $f(t)$ must be zero when $t<0$. This means that $\gamma$ must be chosen so as to be to the right of the pole with the largest real part. That way, when the contour closes to the right, the integral over the closed contour will be zero (as it encloses no poles) and, hence, the ILT is zero by design. Note that it doesn't matter what $\gamma$ is, so long as there are no poles to the right of the line $\Re{z}=\gamma$.
In the simple cases we find from, for example, solving ordinary differential equations with constant coefficients, we will have $\hat{f}(s)$ being a rational function of $s$. In such cases, we will simply have
$$\frac{1}{i 2 \pi} \int_{\gamma-i \infty}^{\gamma+I \infty} ds \, \hat{f}(s) \, e^{s t} = \sum_k \text{Res}_{s=s_k} [\hat{f}(s) \, e^{s t}]$$
As long as you know how to evaluate these residues, then evaluating such ILTs is straightforward. Let's look at a concrete example: solve the differential equation
$$x'' + 4 x= 1$$
with $x(0) = x'(0) = 0$. We apply LTs to both sides and we get
$$\hat{x}(s) = \frac{1}{s (s^2+4)}$$
To obtain $x(t)$, we sum over the residues of $\hat{x}(s) e^{s t}$ at the poles at $s=0$, $s=2 i$, and $s=-2 i$. At $s=0$, the residue is
$$\lim_{s \to 0} s \frac{e^{s t}}{s (s^2+4)} = \frac14$$
The other residues are found similarly:
$$\lim_{s \to 2 i} (s-2 i) \frac{e^{s t}}{s (s^2+4)} = \frac{e^{i 2 t}}{-8}$$
$$\lim_{s \to -2 i} (s+2 i) \frac{e^{s t}}{s (s^2+4)} = \frac{e^{-i 2 t}}{-8}$$
The ILT is the sum of the residues, or
$$x(t) = \frac{1-\cos{2 t}}{4} = \frac12 \sin^2{t}$$
There are times, though, when you need other techniques of contour integration. For example, sometimes you may encounter a LT having a branch point in $s$. One example I have encountered here is $\hat{f}(s) = e^{-\sqrt{s}}$. Here is my evaluation of this ILT. Note how I treated the contour. Despite the fact that there are no poles and hence no residues, the multivaluedness of the LT provides the ILT I sought. I have many more worked examples here.
I hope this helps. Please feel free to ask questions about this topic anytime.