Answer
$$u=e^{\frac{1}{2}-\frac{1}{2}\cos{v^2}}$$
Work Step by Step
$\frac{du}{dv}=uv\sin{v^2}$
$\frac{1}{u}du=v\sin{v^2}dv$
$\int\frac{1}{u}du=\int v\sin{v^2}dv$
$\ln{u}=-\frac{1}{2}\cos{v^2}+C$
$e^{\ln{u}}=u=e^{-\frac{1}{2}\cos{v^2}+C}=e^Ce^{-\frac{1}{2}\cos{v^2}}$
Let $u=e^C$ and we get
$u=ke^{-\frac{1}{2}\cos{v^2}}$.
Now applying our initial condition we get
$1=ke^{-\frac{1}{2}\cos{0^2}}=ke^{-\frac{1}{2}}$
$k=\frac{1}{e^{-\frac{1}{2}}}=e^\frac{1}{2}$
So our particular solution is
$u=e^\frac{1}{2}e^{-\frac{1}{2}\cos{v^2}}=e^{\frac{1}{2}-\frac{1}{2}\cos{v^2}}$
$$u=e^{\frac{1}{2}-\frac{1}{2}\cos{v^2}}$$
