Find u(x,t) with u_(xt)=e^(-t)cos x , u(x,0)=0 , u_t(0,t)=0
Find $u(x,t)$ with $u_{xt}=e^{-t}\cos x$, $u(x,0)=0$, $u_t(0,t)=0$.
Explanation of Method
Integrate the mixed partial once in $x$, fix the resulting arbitrary function of $t$ using $u_t(0,t)=0$, then integrate in $t$ and fix the remaining arbitrary function of $x$ using $u(x,0)=0$.
Step 1 · Integrate in $x$
\[ u_t = e^{-t}\sin x + g(t). \] Setting $x=0$: $g(t)=u_t(0,t)-0=0$, so $u_t=e^{-t}\sin x$.
Step 2 · Integrate in $t$
\[ u = -e^{-t}\sin x + h(x). \] Setting $t=0$: $-\sin x+h(x)=u(x,0)=0 \Rightarrow h(x)=\sin x$.
Verification
$u=\sin x(1-e^{-t})$: $u_t=e^{-t}\sin x$, $u_{xt}=e^{-t}\cos x$. \checkmark $u(x,0)=0$. \checkmark $u_t(0,t)=e^{-t}\sin0=0$. \checkmark
On the stated asymptotic claim
The source problem asks to show $u\to\sin t$ as $x\to\infty$; but $u=\sin x(1-e^{-t})$ has no limit as $x\to\infty$ for fixed $t$ (it oscillates via $\sin x$) and never produces $\sin t$ from $x$ alone. What genuinely holds is the $t$-asymptotic: $u(x,t)\to\sin x$ as $t\to\infty$, since $1-e^{-t}\to1$. The "$\to\sin t$" claim in the original statement appears to be a mismatch (most likely $x$ and $t$ were meant to be swapped somewhere in the source) and is not something this solution satisfies as literally stated.
Result
The solution is: