First write it correctly:
[tex]\int 9 sin(\sqrt{t+1}) dt[/tex]
Make the obvious substitution [itex]u= \sqrt{t+1}= (t+1)^{1/2}[/itex] so that [itex]du= (1/2)(t+1)^{-1/2}dt[/itex] or 2udu= dt.
(Equivalently, u^{2}= t+ 1.)
That changes your integral to
[tex]9\int u sin(u)du[/tex]
and you can integrate that by parts.