evaluate the indefinite integral.\n int e^{3x}sin(3x)dx \n+ c

Explanation:

Step1: Use integration - by - parts

Let $u = \sin(3x)$ and $dv=e^{3x}dx$. Then $du = 3\cos(3x)dx$ and $v=\frac{1}{3}e^{3x}$. By the integration - by - parts formula $\int u;dv=uv-\int v;du$, we have: $\int e^{3x}\sin(3x)dx=\frac{1}{3}e^{3x}\sin(3x)-\int\frac{1}{3}e^{3x}\cdot3\cos(3x)dx=\frac{1}{3}e^{3x}\sin(3x)-\int e^{3x}\cos(3x)dx$

Step2: Use integration - by - parts again on $\int e^{3x}\cos(3x)dx$

Let $u = \cos(3x)$ and $dv = e^{3x}dx$. Then $du=- 3\sin(3x)dx$ and $v=\frac{1}{3}e^{3x}$. $\int e^{3x}\cos(3x)dx=\frac{1}{3}e^{3x}\cos(3x)+\int\frac{1}{3}e^{3x}\cdot3\sin(3x)dx=\frac{1}{3}e^{3x}\cos(3x)+\int e^{3x}\sin(3x)dx$

Step3: Substitute the result of Step2 into Step1

$\int e^{3x}\sin(3x)dx=\frac{1}{3}e^{3x}\sin(3x)-\left(\frac{1}{3}e^{3x}\cos(3x)+\int e^{3x}\sin(3x)dx\right)$ $\int e^{3x}\sin(3x)dx=\frac{1}{3}e^{3x}\sin(3x)-\frac{1}{3}e^{3x}\cos(3x)-\int e^{3x}\sin(3x)dx$ Add $\int e^{3x}\sin(3x)dx$ to both sides: $2\int e^{3x}\sin(3x)dx=\frac{1}{3}e^{3x}\sin(3x)-\frac{1}{3}e^{3x}\cos(3x)$ $\int e^{3x}\sin(3x)dx=\frac{1}{6}e^{3x}(\sin(3x)-\cos(3x))+C$

Answer:

$\frac{1}{6}e^{3x}(\sin(3x)-\cos(3x))$