use integration by parts to evaluate ∫ 3x cos(3x) dx.

Explanation:

Step1: Choose u and dv

Let $u = 3x$, $dv=\cos(3x)dx$.

Step2: Find du and v

Differentiate $u$ to get $du = 3dx$. Integrate $dv$: $\int\cos(3x)dx=\frac{1}{3}\sin(3x)$, so $v=\frac{1}{3}\sin(3x)$.

Step3: Apply integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Substitute $u$, $v$, $du$ into the formula: [ \begin{align*} \int 3x\cos(3x)dx&=3x\times\frac{1}{3}\sin(3x)-\int\frac{1}{3}\sin(3x)\times3dx\ &=x\sin(3x)-\int\sin(3x)dx \end{align*} ]

Step4: Integrate the remaining integral

$\int\sin(3x)dx=-\frac{1}{3}\cos(3x)+C$. So $\int 3x\cos(3x)dx=x\sin(3x)+\frac{1}{3}\cos(3x)+C$.

Answer:

$x\sin(3x)+\frac{1}{3}\cos(3x)+C$