We start by separating variables, and assume $u(x,t) = X(x)T(t).$ Our partial differential equation then becomes $(XT)_{tt} = (9XT)_{xx}$ or $XT'' = 9X''T." Rearranging,
$$\frac{T''}{9T} = \frac{X''}{X} = -\lambda.$$
We have solved the equation $X'' = -\lambda X$ , where $X(0) = X(L) = 0$ many times before, so we know the solution is
The two middle integrals can be evaluated by parts, where we set $u=x,$ $du=dx,$ $v=-\frac{2}{n\pi}\cos\left(\frac{n\pi x}{2}\right),$ and $dv = \sin\left(\frac{n\pi x}{2}\right).$
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We start by separating variables, and assume $u(x,t) = X(x)T(t).$ Our partial differential equation then becomes $(XT)_{tt} = (9XT)_{xx}$ or $XT'' = 9X''T." Rearranging,
$$\frac{T''}{9T} = \frac{X''}{X} = -\lambda.$$
We have solved the equation $X'' = -\lambda X$ , where $X(0) = X(L) = 0$ many times before, so we know the solution is
$$X(x) = \sin\left(\frac{n\pi x}{L}\right) = \sin\left(\frac{n\pi x}{2}\right).$$
(We leave out the constant coefficient for now and have it be "absorbed into" the constant coefficients for $T(t).$)
The solution to $T'' = -9\lambda T$ is
$$T(t) = a\cos(\frac{9\lambda}t) + b\sin(9\sqrt{\lambda}t).$$
Thanks to our solution for $X(x),$ we know $\sqrt{\lambda} = \frac{n\pi}{2},$ so
$$T(t) = a\cos\left(\frac{9n\pi t}{2}\right) + b\sin\left(\frac{9n\pi t}{2}\right).$$
Therefore, the full solution is the series
$$u(x,t) = \sum_{n=1}^{\infty}\left(a_n\cos\left(\frac{9n\pi t}{2}\right) + b_n\sin\left(\frac{9n\pi t}{2}\right)\right)\sin\left(\frac{n\pi x}{2}\right).$$
From our initial conditions, we know
$$u(x,0) = \sum_{n=1}^{\infty}a_n\sin\left(\frac{n\pi x}{2}\right) = 0.$$
Therefore, $a_n = 0,$ so
$$u(x,t) = \sum_{n=1}^{\infty}b_n\sin\left(\frac{9n\pi t}{2}\right)\sin\left(\frac{n\pi x}{2}\right).$$
We also know
$$u_t(x,0) = \frac{9\pi}{2}\sum_{n=1}^{\infty}nb_n\sin\left(\frac{n\pi x}{2}\right) = -|x-1|,$$
so
$$\sum_{n=1}^{\infty}nb_n\sin\left(\frac{n\pi x}{2}\right) = -\frac{2}{9\pi}|x-1|.$$
From our study of Fourier series, we know
$$b_n = \frac{2}{2}\int_{0}^{2}-\frac{2}{9n\pi}|x-1|\sin\left(\frac{n\pi x}{2}\right)dx = -\frac{2}{9n\pi}\int_{0}^{2}|x-1|\sin\left(\frac{n\pi x}{2}\right)dx$$
$$= -\frac{2}{9n\pi}(\int_{0}^{1}\left(1-x)\sin\left(\frac{n\pi x}{2}\right)dx + \int_{1}^{2}(x-1)\sin\left(\frac{n\pi x}{2}\right)dx\right)$$
$$= -\frac{2}{9n\pi}\left(\int_{0}^{1}\sin\left(\frac{n\pi x}{2}\right)dx + \int_{0}^{1}x\sin\left(\frac{n\pi x}{2}\right)dx + \int_{1}^{2}x\sin\left(\frac{n\pi x}{2}\right)dx + \int_{1}^{2}\sin\left(\frac{n\pi x}{2}\right)dx\right).$$
The two middle integrals can be evaluated by parts, where we set $u=x,$ $du=dx,$ $v=-\frac{2}{n\pi}\cos\left(\frac{n\pi x}{2}\right),$ and $dv = \sin\left(\frac{n\pi x}{2}\right).$
$$b_n = \frac{2}{9n\pi}\frac{2}{n\pi}\left(\left[\cos\left(\frac{n\pi x}{2}\right)\right]_0^1 - \left(\left[x\cos\left(\frac{n\pi x}{2}\right)\right]_0^1 - \int_{0}^{1}\cos\left(\frac{n\pi x}{2}\right)dx\right) + \left(\left[x\cos\left(\frac{n\pi x}{2}\right)\right]_1^2 - \int_{1}^{2}\cos\left(\frac{n\pi x}{2}\right)dx\right) - \left[\cos\left(\frac{n\pi x}{2}\right)\right]_1^2\right)$$
$$ = \frac{4}{9n^2\pi^2}(((\cos(\frac{n\pi}{2})-1) - (\cos(\frac{n\pi}{2}) - \frac{2}{n\pi}\left[\sin(\frac{n\pi x}{2}\right]_0^1) + ((2\cos(n\pi)-\cos(\frac{n\pi}{2})) - \frac{2}{n\pi}\left[\sin(\frac{n\pi x}{2})\right]_1^2) - (\cos(n\pi)-\cos\frac{n\pi}{2}))).$$
Several of the terms above cancel out, leaving us with
$$\frac{4}{9n^2\pi^2}\left(-1 + \cos(n\pi) + \frac{2}{n\pi}\left(\left[\sin\left(\frac{n\pi x}{2}\right)\right]_0^1 - \left[\sin\left(\frac{n\pi x}{2}\right)\right]_1^2\right)\right)$$
$$= \frac{4}{9n^2\pi^2}\left(-1 + \cos(n\pi) + \frac{2}{n\pi}\left(\sin\left(\frac{n\pi}{2}\right) - \left(-\sin\left(\frac{n\pi}{2}\right)\right)\right)\right) = \frac{4}{9n^2\pi^2}\left(\cos(n\pi)-1 + \frac{4}{n\pi}\sin\left(\frac{n\pi}{2}\right)\right)$$
$$= \frac{4}{9n^2\pi^2}\left((-1)^n -1 + \frac{4}{n\pi}\sin\left(\frac{n\pi}{2}\right)\right).$$
Thus,
$$u(x,t) = \frac{4}{9\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}\left((-1)^n-1+\frac{4}{n\pi}\sin\left(\frac{n\pi}{2}\right)\right)\sin\left(\frac{9n\pi t}{2}\right)\left(\frac{n\pi x}{2}\right).$$