make a substitution and then use integration by parts to evaluate ∫ 3 cos(√x) dx.

Explanation:

Step1: Make a substitution

Let $u = \sqrt{x}$, then $x = u^{2}$ and $dx=2u\ du$. The integral $\int 3\cos(\sqrt{x})dx$ becomes $\int 3\cos(u)\cdot 2u\ du=6\int u\cos(u)du$.

Step2: Apply integration - by - parts formula

The integration - by - parts formula is $\int vdw=vw-\int wdv$. Let $v = u$ and $dw=\cos(u)du$. Then $dv = du$ and $w=\sin(u)$. So, $6\int u\cos(u)du=6\left[u\sin(u)-\int\sin(u)du\right]$.

Step3: Integrate $\sin(u)$

We know that $\int\sin(u)du=-\cos(u)+C$. So, $6\left[u\sin(u)-\int\sin(u)du\right]=6\left[u\sin(u)+\cos(u)\right]+C$.

Step4: Substitute back $u = \sqrt{x}$

Substituting $u=\sqrt{x}$ back into the expression, we get $6\left[\sqrt{x}\sin(\sqrt{x})+\cos(\sqrt{x})\right]+C$.

Answer:

$6\sqrt{x}\sin(\sqrt{x}) + 6\cos(\sqrt{x})+C$