separable solution for the wave equation given classical wave equation:
$$
\frac{\partial^{2} u}{\partial t^{2}}-b^{2} \frac{\partial^{2} u}{\partial x^{2}}=0, \quad 0 \leq x \leq 1 .
$$
given also boundary conditions:
$$
u(0, t)=u(1, t)=0, \quad t>0
$$
and initial conditions:
$$
u(x, 0)=0.5 x(1-x), \quad \frac{\partial u}{\partial t}(x, 0)=0 \quad 0 \leq x \leq 1
$$
I want to show that analytical solution of the problem with b = 1 is:
$$
u(x, t)=\frac{2}{\pi^{3}} \sum_{n=1}^{\infty} \frac{1}{n^{3}}[1-\cos (\pi n)] \cos (\pi n t) \sin (\pi n x)
$$
I don't understand, I want to use the integral, I don't understand where the use of the integral can help me
 A: Since you already have the solution, in order to verify that at given $u(x,t)$ is indeed a solution, you have to individually verify that it satisfies :

*

*the differential equation

*the boundary conditions

*the initial conditions

1) The differential equation
\begin{align}
\frac{\partial^2 u(x,t)}{\partial{t}^2} &= \frac{2}{\pi^{3}} \sum_{n=1}^{\infty} \frac{1}{n^{3}}[1-\cos (\pi n)](-\pi^2 n^2) \cos (\pi n t) \sin (\pi n x) \\
\frac{\partial^2 u(x,t)}{\partial{x}^2} &= \frac{2}{\pi^{3}} \sum_{n=1}^{\infty} \frac{1}{n^{3}}[1-\cos (\pi n)](-\pi^2 n^2) \cos (\pi n t) \sin (\pi n x) 
\end{align}
Therefore,
$$ \frac{\partial^2 u(x,t)}{\partial{t}^2} - \frac{\partial^2 u(x,t)}{\partial{x}^2} = 0$$
2) The boundary conditions
\begin{align}
u(0,t) &= \frac{2}{\pi^{3}} \sum_{n=1}^{\infty} \frac{1}{n^{3}}[1-\cos (\pi n)]\cos (\pi n t) \sin (0) = 0 \\
u(1,t) &= \frac{2}{\pi^{3}} \sum_{n=1}^{\infty} \frac{1}{n^{3}}[1-\cos (\pi n)] \cos (\pi n t) \sin (\pi n) = 0
\end{align}
Therefore,
$$u(0,t) = u(1,t) = 0$$
3) The initial conditions
\begin{align}
\frac{\partial u}{\partial{t}}(x,0) = &= \frac{2}{\pi^{3}} \sum_{n=1}^{\infty} \frac{1}{n^{3}}[1-\cos (\pi n)](-\pi n )\sin (0) \sin (\pi n x) = 0
\end{align}
The tricky part consist of demonstrating that, for $0 < x < 1$,
$$u(x,0) = \frac{2}{\pi^{3}} \sum_{n=1}^{\infty} \frac{1}{n^{3}}[1-\cos (\pi n)]\sin (\pi n x) = \frac{1}{2}x(1-x)$$
This can be demonstrated by verifying that their first and second derivatives of both sides are indeed equal
\begin{align}
\frac{\partial{u}}{\partial{x}}(x,0) = \frac{2}{\pi^{2}} &\sum_{n=1}^{\infty} \frac{1}{n^{2}}[1-\cos (\pi n)]\cos (\pi n x) = \frac{1}{2} - x \\
-\frac{\partial^2{u}}{\partial{x}^2}(x,0) = \frac{2}{\pi} &\sum_{n=1}^{\infty} \frac{1}{n}[1-\cos (\pi n)]\sin(\pi n x) = 1 
\end{align}
The second one is equivalent to show that,
$$\sum_{n=1}^{\infty} \frac{1}{2n+1}\sin(\pi (2n+1) x) = \frac{\pi}{4}$$
which by the way, a plot shows us, that is only true for $0 < x < 1$. This infinite Fourier series is well known, and you can find a demonstration of it here.
That the second derivatives are equal does not imply by itself that that the function are equal, you also have to verify that the function and its first derivative are equal in one point, which you should be able to do.
