Answer
$$\int\frac{\sin\sqrt x}{\sqrt x}dx=-2\cos\sqrt x+C$$
Work Step by Step
$$A=\int\frac{\sin\sqrt x}{\sqrt x}dx$$
Let $u=\sqrt x=x^{1/2}$.
Then $du=\frac{1}{2}x^{-1/2}dx=\frac{1}{2\sqrt x}dx$, so $\frac{1}{\sqrt x}dx=2du$
Substitute into $A$, we have $$A=2\int\sin udu$$ $$A=-2\cos u+C$$ $$A=-2\cos\sqrt x+C$$
