Hey everyone. I'm having trouble understanding this example. It says,

Find the inverse of $\displaystyle H(s) = \frac{1}{(s^2+w^2)^2}$

$\displaystyle F(s) = \frac{1}{s^2+w^2}$, so $\displaystyle f(t) = \sin{wt}/w$

$\displaystyle \frac{\sin{wt}}{w} * \frac{\sin{wt}}{w} = \frac{1}{w^2} \int^t_0 \sin{wu}\sin{w(t-u)}du$

So far everything is straightforward. Then the solution says:

From the trigonometric formula $\displaystyle \sin{x}\sin{y} = \frac{1}{2}[-\cos{(x+y)} + cos{(x-y)}]$ with $\displaystyle x = \frac{1}{2}(wt + wu)$ and $\displaystyle y = \frac{1}{2}(wt-wu)$, we obtain

$\displaystyle \frac{1}{2w^2}\int^t_0 [-\cos{wt} + \cos{wu}]du$

Okay. If I plug x and y into the given trig formula, it comes out as the integral above. However,
$\displaystyle \sin{wu} \neq \sin{\frac{1}{2}(wt+wu)}$
$\displaystyle \sin{w(t-u)} \neq \sin{\frac{1}{2}(wt-wu)}$

I'm confused to how/why they did that because it doesn't make sense (at least not to me).
Any help is appreciated. Thanks in advance!