Let $H(s) = \frac{1}{(s^2 + w^2)^2}$
Then $\displaystyle h(t) = \frac{\sin(wt)}w * \frac{\sin(wt)}w = \frac 1{w^2} \int_0^t \sin(w \tau) \sin(w(t-\tau)) \,d\tau$
$\displaystyle = \frac 1{2w^2} \int_0^t -\cos(wt) + \cos(w\tau) \,d\tau$
How do you get this from the trig identity $\sin(x)\sin(y) = \frac12 ( -\cos(x+y) + \cos(x-y))$ or is it a typo in the book? I get the following:
$$h(t) = \frac1{2w^2} \int_0^t -\cos(w\tau) + \cos(2w\tau - wt)\, d\tau$$
Let's start from the top.
$$F(s) = \frac1{(s^2+w^2)^2} $$
Then the ILT is
$$\begin{align}f(t) = \frac{\sin{w t}}{w t} * \frac{\sin{w t}}{w t} &= \frac1{w^2} \int_0^t d\tau \sin{w \tau} \sin{w (t-\tau)}\\ &= \frac1{2 w^2} \int_0^t d\tau \left [\cos{w(2 \tau - t)} - \cos{w t} \right ] \\ &=\frac1{2 w^3} \left [\sin{w t}- w t \cos{w t} \right ]\end{align}$$
So far your answer looks correct, but let's check by actually evaluating the integral definition of the ILT, i.e., the sum of the residues at the poles $\pm i w$:
$$\begin{align}f(t) &= \left [\frac{d}{dz} \frac{e^{z t}}{(z+i w)^2} \right ]_{z=i w} + \left [\frac{d}{dz} \frac{e^{z t}}{(z-i w)^2} \right ]_{z=-i w}\\ &= -2 (i 2 w)^{-3} e^{i w t} + (i 2 w)^{-2} t e^{i w t} - 2 (-i 2 w)^{-3} e^{-i w t} + (-i 2 w)^{-2} t e^{-i w t} \\ &=\frac{\sin{w t}}{2 w^3} - \frac{t \cos{w t}}{2 w^2}\end{align}$$
which agrees with the above result.
It urns out, however, that the book's result is correct as well. You can check this by carrying out the integration. It is not clear to me, however, how this result was derived.
