question evaluate the integral below. ∫e^2x sin(3x) dx be sure to place the argument of any trigonometric or…

Explanation:

Step1: Use integration - by - parts formula $\int u;dv=uv-\int v;du$

Let $u = \sin(3x)$ and $dv=e^{2x}dx$. Then $du = 3\cos(3x)dx$ and $v=\frac{1}{2}e^{2x}$. So, $\int e^{2x}\sin(3x)dx=\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{2}\int e^{2x}\cos(3x)dx$.

Step2: Apply integration - by - parts again on $\int e^{2x}\cos(3x)dx$

Let $u=\cos(3x)$ and $dv = e^{2x}dx$. Then $du=- 3\sin(3x)dx$ and $v=\frac{1}{2}e^{2x}$. So, $\int e^{2x}\cos(3x)dx=\frac{1}{2}e^{2x}\cos(3x)+\frac{3}{2}\int e^{2x}\sin(3x)dx$.

Step3: Substitute the result of $\int e^{2x}\cos(3x)dx$ into the first integration - by - parts result

$\int e^{2x}\sin(3x)dx=\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{2}\left(\frac{1}{2}e^{2x}\cos(3x)+\frac{3}{2}\int e^{2x}\sin(3x)dx\right)$. $\int e^{2x}\sin(3x)dx=\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{4}e^{2x}\cos(3x)-\frac{9}{4}\int e^{2x}\sin(3x)dx$.

Step4: Solve for $\int e^{2x}\sin(3x)dx$

Add $\frac{9}{4}\int e^{2x}\sin(3x)dx$ to both sides: $\left(1 + \frac{9}{4}\right)\int e^{2x}\sin(3x)dx=\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{4}e^{2x}\cos(3x)$. $\frac{13}{4}\int e^{2x}\sin(3x)dx=\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{4}e^{2x}\cos(3x)$. $\int e^{2x}\sin(3x)dx=\frac{2}{13}e^{2x}\sin(3x)-\frac{3}{13}e^{2x}\cos(3x)+C$.

Answer:

$\frac{2}{13}e^{2x}\sin(3x)-\frac{3}{13}e^{2x}\cos(3x)+C$