Fourier's method for PDE We have $$U_{tt}=U_{xx} \quad 0<x<{\pi \over 2}, \quad t>0$$
$$U(x,0)=U(0,t)=U_x({\pi \over 2},t)=0$$
$$U_t(x,0)=\cos(5x) \cdot \sin(x)$$
We are looking for a solution in the form $U(x,t)=X(x)T(t)$
Then $U_x=X'(x)T(t) \quad U_t=X(x)T'(t)$
and $U_{xx}=X''(x)T(t) \quad U_{tt}=X(x)T''(t)$
From what is given $X''(x)T(t)=X(x)T''(t)\Rightarrow \exists \lambda=\text{const}:{X''(x) \over X(x)}={T''(t) \over T(t)}=\lambda$
Now $U(0,t)=0=X(0)T(t)$ but $U \not =0 \Rightarrow T \not = 0 \Rightarrow X(0)=0$
And $U_x({\pi \over 2},t)=0=X'({\pi \over 2})T(t)$ but $U \not =0 \Rightarrow T \not = 0 \Rightarrow X'({\pi \over 2})=0$
We thus got the Sturm–Liouville problem ${X''(x) \over X(x)}=\lambda \land X(0)=0=X'({\pi\over2})$
We proceed to find $\lambda=-(2k+1)^2 \land X(x)=C_k\cdot \sin[(2k+1)x]$
We now examine ${T''(t) \over T(t)}=\lambda=-(2k+1)^2$
We get $T(t)=\hat A_k \cdot \cos (2k+1)t+\hat B_k \cdot \sin(2k+1)t$
$\Rightarrow U(x,t)=X(x)T(t)=\sin(2k+1)x[ A_k \cdot \cos (2k+1)t+B_k \cdot \sin(2k+1)t]$
Where $A_k=\hat A_kC_k \land B_k=\hat B_kC_k$
We are looking for a solution in the form $$U(x,t)=\sum_{k=0}^\infty \sin(2k+1)x[ A_k \cdot \cos (2k+1)t+B_k \cdot \sin(2k+1)t]$$
We have $U(x,0)=0=\sum_{k=0}^\infty \sin(2k+1)x\cdot A_k$
As we know the scalar product in $L_2$ we get $$A_k \int_0^{\pi \over 2} {\sin^2[(2k+1)x]\,\mathrm{d}x}=0$$
Let us examine $$\int_0^{\pi \over 2} {\sin^2[(2k+1)x]\,\mathrm{d}x} = \int_0^{\pi \over 2} {{1-\cos[(4k+2)x]\over 2}\,\mathrm{d}x}={\pi \over 4}-{{1 \over 2}\int_0^{\pi \over 2}} {\cos[(4k+2)x]\,\mathrm{d}x}={\pi \over 4}-{1 \over 2(4k-2)}\int_0^{\pi \over 2} {\cos[(4k+2)x]\,\mathrm{d}[(4k+2)x]}={\pi \over 4}-{1 \over 2(4k-2)}\int_0^{\pi \over 2} {\cos(x)\,\mathrm{d}x}={\pi \over 4}-{1 \over 8k-4}[\sin({\pi\over 2})-\sin(0)]={\pi \over 4}-{1 \over 8k-4}$$
This is my calculation. Note that $$\int {\cos[(4k+2)x]\,\mathrm{d}x}={1 \over 4k-2}\int {\cos[(4k+2)x]\,\mathrm{d}[(4k+2)x]}={1 \over 4k-2}\int {\cos(x)\,\mathrm{d}x}={1 \over 4k-2}\sin(x)$$ as $$\int {f(x)\,\mathrm{d}x}={1 \over a}\int {f(x)\,\mathrm{d}(ax)}$$
Here is where wolfram-alpha disagrees. They say that $$\int_0^{\pi \over 2} {\cos[(4k+2)x]\,\mathrm{d}x}=0$$ since $$\int {\cos(ax)\,\mathrm{d}x}={1 \over a} \sin(ax)$$
Either way all $A_k=0$. But to get $B_k$ we use $$U_t(x,0)=\cos(5x)\sin(x)=\sum_{k=0}^\infty \sin(2k+1)x\cdot B_k \cdot (2k+1)$$
$$\Rightarrow (2k+1)B_k\int_0^{\pi \over 2} {\sin^2[(2k+1)x]\,\mathrm{d}x} = \int_0^{\pi \over 2} \sin[(2k+1)x]\cos(5x){\sin(x)\,\mathrm{d}x}$$
1) Can you help me with those two integrals.
2) Do I need to verify that the series converges, and how? Which criterion should I use? Does it not follow from the general theory of these equations that the series converges? When does it?
 A: $$X'' + \lambda^2 X = 0$$
subject to $X(0) = X'(\pi/2) = 0$.  When you write the solution as
$$X(x) = A \cos{\lambda x} + B \sin{\lambda x}$$
you see that the condition at $x=0 \implies A=0$.  The condition at $x=\pi/2$ implies that, for $k \in \mathbb{Z}$: 
$$\frac{\pi}{2} \lambda = (2 k+1) \pi/2 \implies \lambda = 2 k + 1 $$
so the spatial solution is 
$$X(x) = B \sin{(2 k+1) x}$$
We also see that $T'' + (2 k+1)^2 T = 0$ subject to $T(0)=0$ and $T'(0) = \cos{5 x} \sin{x} = (\sin{6 x} - \sin{4 x})/2$
So you are looking for a solution of the form
$$U(x,t) = \sum_k B_k \sin{(2 k+1) x} \sin{(2 k+1) t}$$
The $B_k$ are found by applying the second initial condition (the derivative at $t=0$):
$$\sum_k ((2 k+1) B_k \sin{(2 k+1) x} = \frac{\sin{6 x} - \sin{4 x}}{2} $$
Because the Fourier series forms an orthogonal basis, we may find that the $B_K$ satisfy
$$\begin{align}(2 k+1) B_k \int_0^{\pi/2} dx \: \sin^2{[(2 k+1) x]} &= \frac{1}{2} \int_0^{\pi/2} dx \:  \sin{[(2 k+1) x]} (\sin{6 x} - \sin{4 x})\\ (2 k+1) B_k \frac{\pi}{4} &= (-1)^{k+1}\left ( \frac{3}{ 4 k^2 +4 k -35} + \frac{2 }{ 4 k^2 +4 k -15} \right )  \end{align}$$
The integrals on the right are accomplished by repeated application of the sine product formula I used above.  The integral on the left may be found by using $\sin^2{[(2 k+1) x]} = (1-\cos{[2 (2 k+1) x]})/2$.
Because the $B_k \sim 1/k^3$ for large $k$, even without the alternating series, the series would converge.
