question evaluate the integral below. ∫ - 3x cos(3x) dx be sure to place the argument of any trigonometric…

question evaluate the integral below. ∫ - 3x cos(3x) dx be sure to place the argument of any trigonometric or logarithmic functions in parentheses in your answer. provide your answer below: ∫ - 3x cos(3x) dx =

question evaluate the integral below. ∫ - 3x cos(3x) dx be sure to place the argument of any trigonometric or logarithmic functions in parentheses in your answer. provide your answer below: ∫ - 3x cos(3x) dx =

Answer

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = - 3x$ and $dv=\cos(3x)dx$. Then $du=-3dx$ and $v=\frac{1}{3}\sin(3x)$.

Step2: Substitute into the formula

$\int - 3x\cos(3x)dx=-3x\times\frac{1}{3}\sin(3x)-\int\frac{1}{3}\sin(3x)\times(-3)dx$. Simplify to get $-x\sin(3x)+\int\sin(3x)dx$.

Step3: Integrate $\sin(3x)$

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

Answer:

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