In section 7.2 of Evans PDE book, he proves that there exists a unique weak solution of \begin{equation*} \left\{ \begin{array}{l} u_{tt}+Lu=f, \quad \textrm{in } U\times (0,T] \\ u=0, \quad \textrm{on } \partial U\times [0,T] \\ u=g, \ u_t=h, \quad \textrm{on } U\times \{t=0\} \\ \end{array} \right. \end{equation*} where $$ Lu= -\sum_{i,j=1}^n a^{ij}(x,t)u_{x_ix_j}+\sum_{i=1}^n b^{i}(x,t)u_{x_i}+c(x,t)u.$$
My problem is in one equality of the uniqueness part of the demonstration. For that, he considers the problem \begin{equation*} \left\{ \begin{array}{l} u_{tt}+Lu=0, \quad \textrm{in } U\times (0,T] \\ u=0, \quad \textrm{on } \partial U\times [0,T] \\ u=0, \ u_t=0, \quad \textrm{on } U\times \{t=0\} \\ \end{array} \right. \end{equation*} He defines \begin{equation*} v:=\left\{ \begin{array}{l} \int_t^s u(\tau)d\tau, \quad \textrm{if } 0\leq t \leq s \\ 0, \quad \textrm{if } s\leq t \leq T. \end{array} \right. \end{equation*}
Since $v\in H_0^1(U)$, we can multiply the first equation of our system by $v$ and integrate from $0$ to $s$ to obtain $$ \int_0^s <u'',v>+B[u,v;t]dt=0$$ where $B[u,v;t]:=\int_U \sum_{i,j=1}^n a^{ij}(x,t)u_{x_i}v_{x_j}+\sum_{i=1}^n b^{i}(x,t)u_{x_i}v+c(x,t)uvdx$.
He uses the fact that $v'=-u$ and $u'(0)=v(s)=0$ to obtain $$ \int_0^s <u',u>-B[v',v;t]dt=0.$$
Up until here, I can understand everything. My problem is in the next equality. He then puts $$ \int_0^s \frac{1}{2}\frac{d}{dt}\left(||u||_{L^2}^2-B[v,v;t]\right) dt=-\int_0^s C[u,v;t]+D[v,v;t]$$ where $$ C[u,v;t]:=\int_U \sum_{i=1}^n b^i v_{x_i}u+\frac{1}{2}b_{i,x_i}uv \ dx$$ and $$ D[u,v;t]:=\int_U \sum_{i,j=1}^n a_{ij,t}u_{x_i} v_{x_j}+\sum_{i=1}^n b_{i,t} u_{x_i}v+c_t uv \ dx.$$
I can see how he gets the left side but how do I get the right side? It is not obvious for me that the equality holds.