Showing the general solution reduces to $x(t)\approx \frac{F}{2mw^2_0\theta}\left(\sin(w_0t)-\sin(wt)\right)$ 
Consider the equation for a periodically forced, damped oscillator:
  $$m\ddot{x}+r_0\dot{x}+k_0x=F\sin(wt)$$
  where $m, r_0, k_0$ are positive constants. Suppose now that $r_0=0$ and let $w_0=\sqrt{\frac{k_0}{m}}$.

I have already shown that the general solution to the above ODE satisfying the inital conditions $x(0)=x_0, \ \dot{x}(0)=v_0$ is 
$$x(t)=x_0\cos(w_0t)+\frac{1}{w_0}\left(v_0-\frac{Fw}{m(w^2_0-w^2)}\right)\sin(w_0t)+\frac{F}{m(w^2_0-w^2)}\sin(wt)$$

Show that is $w\approx w_0$, say if $w=w_0(1+\theta)$ where $|\theta|<<1$, then
  $$x(t)\approx \frac{F}{2mw^2_0\theta}\left(\sin(w_0t)-\sin(wt)\right)$$

I can't show this result. Substituting $w=w_0(1+\theta)$ and using simple algebra, I am stuck at
$$x(t)=x_0\cos(w_0t)+\sin(w_0t)\left(\frac{v_0}{w_o}+\frac{F}{mw^2_0(2+\theta)}\right)+\frac{F}{mw^2_0\theta(2+\theta)}\left(\sin(w_0t)-\sin(wt)\right)$$
It appears this is just an exercise in algebra, but I'm so stuck. Any advice would be really appreciated.
 A: I shall put my faith in your calculations and believe that
$$x(t)=x_0\cos(w_0t)+\frac{1}{w_0}\left(v_0-\frac{Fw}{m(w^2_0-w^2)}\right)\sin(w_0t)+\frac{F}{m(w^2_0-w^2)}\sin(wt)$$
is true.  Note that we can ignore every term except those with denominator $w_0^2-w^2$ (since this number will be almost $0$, making the terms with this denominator blow up).  Hence,
$$x(t)\approx-\frac{Fw}{mw_0(w_0^2-w^2)}\,\sin(w_0t)+\frac{F}{m(w_0^2-w)}\,\sin(wt)\,.$$
Consequently, note that $w_0^2-w^2=w_0^2-w_0^2(1+\theta)^2\approx -2\theta w_0^2$, where terms with second or higher orders of $\theta$ are neglected.  Thus,
$$x(t)\approx \frac{Fw}{2mw_0^3\theta}\,\sin(w_0t)-\frac{F}{2mw_0^2\theta}\,\sin(wt)\,.\tag{*}$$
As $\dfrac{w}{w_0}\approx 1$, we obtain
$$x(t)\approx \frac{F}{2mw_0^2\theta}\,\sin(w_0t)-\frac{F}{2mw_0^2\theta}\,\sin(wt)\,,\tag{#}$$
which is what we need.

Caveat:  This estimation is not entirely correct, and I personally would not use it, at least in the small-$t$ limit (namely, $|t|\ll\dfrac{\pi}{2w_0\theta}$).  First, remember when we used the approximation $\dfrac{w}{w_0} \approx 1$.  In fact, we have $\dfrac{w}{w_0}=1+\theta$.  Thus, already, (*) should become
$$x(t)\approx \frac{F}{2mw_0^2}\,\sin(w_0t)+\frac{F}{2mw_0}\,\left(\frac{\sin(w_0t)-\sin(wt)}{w_0\theta}\right)\,.$$
For small $t$, we have
$$\frac{\sin(w_0t)-\sin(wt)}{w_0\theta t}\approx-\cos(w_0t)\,,$$
whence
$$x(t)\approx \frac{F}{2mw_0^2}\,\sin(w_0t)-\frac{Ft}{2mw_0}\,\cos(w_0t)$$
is a better approximation, but then from the $\theta$-independence of this result, the assumption that "we can ignore every term except those with denominator $w_0^2-w^2$ [...]" is no longer valid.  Indeed, we need every other term, and here is a much more correct estimate for small $t$:
$$x(t)\approx x_0\,\cos(w_0t)+\frac{v_0}{w_0}\,\sin(w_0 t)+\frac{F}{2mw_0^2}\,\sin(w_0t)-\frac{Ft}{2mw_0}\,\cos(w_0t)\,.$$
However, for large $t$, the values of $\sin(w_0t)$ and $\sin(wt)$ can be sufficiently large.  Ergo, the approximation (#) is only good for large values of $t$.  In fact, (#) is not even good for all large $t$.  Let $n$ denote the integer closest to $\dfrac{w_0\theta t}{\pi}$ (if there are two of such integers, then you can choose any of them to be $n$).  In the case that $\left|t-\dfrac{\pi}{w_0\theta}n\right|\ll\dfrac{\pi}{2w_0\theta}$, a better approximation is
$$x(t)\approx x_0\,\cos(w_0t)+\frac{v_0}{w_0}\,\sin(w_0 t)+\frac{F}{2mw_0^2}\,\sin(w_0t)-\frac{F}{2mw_0}\,\left(t-\frac{\pi}{w_0\theta}n\right)\,\cos(w_0t)\,.$$
If $\left|t-\dfrac{\pi}{w_0\theta}n\right|$ is approximately in the same order as $\dfrac{\pi}{2w_0\theta}$, then the estimate (#) is good.
