Answer
$$\int\frac{1}{\sqrt t}\cos\Big(\sqrt t+3\Big)dt=2\sin(\sqrt t+3)+C$$
Work Step by Step
$$A=\int\frac{1}{\sqrt t}\cos\Big(\sqrt t+3\Big)dt$$
We set $u=\sqrt t+3$
Then $$du=\frac{1}{2\sqrt t}dt$$
That means $$\frac{1}{\sqrt t}dt=2du$$
Therefore, $$A=2\int\cos udu=2\sin u+C$$ $$A=2\sin(\sqrt t+3)+C$$
