Answer
\[2\left[\sqrt{t}\sin\sqrt{t}+\cos\sqrt{t}\right]+C\]
Work Step by Step
Let \[I=\int\cos\sqrt{t} \:dt\]
Substitute $\:x=\sqrt{t}\;\;\;\ldots (1)$
\[\Rightarrow dx=\frac{1}{2\sqrt{t}}dt\]
\[I=2\int x\cos x \;dx\]
Using integration by parts
\[I=2\left[x\int\cos x \;dx-\int\left((x)'\int\cos x \;dx\right)\;dx\right]+C\]
Where $C$ is constant of integration
\[I=2\left[x\sin x-\int\sin x\:dx\right]+C\]
\[I=2\left[x\sin x+\cos x\right]+C\]
From (1)
\[I=2\left[\sqrt{t}\sin\sqrt{t}+\cos\sqrt{t}\right]+C\]
Hence ,
\[I=2\left[\sqrt{t}\sin\sqrt{t}+\cos\sqrt{t}\right]+C.\]
