Laplace transforms with Fresnel(?) integrals

Hint: $\cos(\tau)$ is the real part of $\exp(i\tau)$, which you combine with the $\exp(-p \tau)$ in one integral, and use the definition of the Gamma function.

So, the table I'm looking in (on Wikipedia) doesn't have an exact match, but it does have several related entries which we can use to build this. (That table uses the letter $s$ instead of $p$, but it's your problem; I'll defer to your notation)

In all that follows, $u(t)$ denotes the unit step function, $1$ for positive $t$ and $0$ for negative $t$. This is equivalent to taking our $t$ integrals from $0$ to $\infty$ instead of on the whole real line.

$$\mathcal{L}\left(t^q\cdot u(t)\right) = \frac{\Gamma(q+1)}{p^{q+1}}$$ $$\mathcal{L}\left(e^{at}f(t)\right) = F(p-a)$$ That's one entry in the table, plus the frequency shift property; $F$ is the Laplace transform of $f$. Yes, it doesn't get us $\sin$ directly, but we can write $\sin$ as a linear combination of complex exponentials $\sin(t) = \frac1{2i}\left(e^{it} - e^{-it}\right)$. Therefore, $$\mathcal{L}\left(\sin(at)f(t)\right) = \frac1{2i}\left(F(p-ai)-F(p+ai)\right)$$ Applied to our transform for the reciprocal square root, that gives $$\mathcal{L}\left(\sin(t)\cdot t^{-\frac12}\right) = \frac{\Gamma(\frac12)}{2i}\left(\frac1{\sqrt{p-i}}-\frac1{\sqrt{p+i}}\right) = \frac{\sqrt{\pi}}{2i}\cdot\frac{\sqrt{p+i}-\sqrt{p-i}}{\sqrt{p^2+1}}$$ $$\mathcal{L}\left(\sin(t)\cdot t^{-\frac12}\right) = \frac{\sqrt{\pi}}{\sqrt{p^2+1}\left(\sqrt{p+i}+\sqrt{p-i}\right)}$$ That is real for real $p$; we just can't get rid of the imaginary numbers in the expression while keeping a (algebraic) closed form.

Handling sines and cosines in the Laplace transform gets a lot easier when you bring in the complex exponential, and allow for complex values of $p$.