$\displaystyle \displaystyle\int\frac{\cos\sqrt{x}}{\sqrt{x}}dx$

Let $\displaystyle u=\sqrt{x}=x^{\frac{1}{2}}$

$\displaystyle du =\frac{1}{2\sqrt{x}} dx$

$\displaystyle \displaystyle\int\frac{\cos u}{u} du$

$\displaystyle \displaystyle\int(\cos u)u^{\frac{-1}{2}} du$

I'm new to integrals and to the substitution rule. I don't know how to integrate products yet (besides by multiplying them out first). I'm kinda stuck right here. Can someone please give me a hint to get me unstuck?

Thanks