question evaluate the integral below. ∫e^5x sin(3x)dx be sure to place the argument of any trigonometric or…

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = \sin(3x)$ and $\mathrm{d}v=e^{5x}\mathrm{d}x$. Then $\mathrm{d}u = 3\cos(3x)\mathrm{d}x$ and $v=\frac{1}{5}e^{5x}$. So, $\int e^{5x}\sin(3x)\mathrm{d}x=\frac{1}{5}e^{5x}\sin(3x)-\frac{3}{5}\int e^{5x}\cos(3x)\mathrm{d}x$.

Step2: Apply integration - by - parts again on $\int e^{5x}\cos(3x)\mathrm{d}x$

Let $u=\cos(3x)$ and $\mathrm{d}v = e^{5x}\mathrm{d}x$. Then $\mathrm{d}u=- 3\sin(3x)\mathrm{d}x$ and $v=\frac{1}{5}e^{5x}$. So, $\int e^{5x}\cos(3x)\mathrm{d}x=\frac{1}{5}e^{5x}\cos(3x)+\frac{3}{5}\int e^{5x}\sin(3x)\mathrm{d}x$.

Step3: Substitute the result of $\int e^{5x}\cos(3x)\mathrm{d}x$ into the first integration - by - parts result

$\int e^{5x}\sin(3x)\mathrm{d}x=\frac{1}{5}e^{5x}\sin(3x)-\frac{3}{5}\left(\frac{1}{5}e^{5x}\cos(3x)+\frac{3}{5}\int e^{5x}\sin(3x)\mathrm{d}x\right)$. Expand the right - hand side: $\int e^{5x}\sin(3x)\mathrm{d}x=\frac{1}{5}e^{5x}\sin(3x)-\frac{3}{25}e^{5x}\cos(3x)-\frac{9}{25}\int e^{5x}\sin(3x)\mathrm{d}x$.

Step4: Solve for $\int e^{5x}\sin(3x)\mathrm{d}x$

Add $\frac{9}{25}\int e^{5x}\sin(3x)\mathrm{d}x$ to both sides: $\int e^{5x}\sin(3x)\mathrm{d}x+\frac{9}{25}\int e^{5x}\sin(3x)\mathrm{d}x=\frac{1}{5}e^{5x}\sin(3x)-\frac{3}{25}e^{5x}\cos(3x)$. $\left(1 + \frac{9}{25}\right)\int e^{5x}\sin(3x)\mathrm{d}x=\frac{1}{5}e^{5x}\sin(3x)-\frac{3}{25}e^{5x}\cos(3x)$. $\frac{34}{25}\int e^{5x}\sin(3x)\mathrm{d}x=\frac{1}{5}e^{5x}\sin(3x)-\frac{3}{25}e^{5x}\cos(3x)$. Multiply both sides by $\frac{25}{34}$: $\int e^{5x}\sin(3x)\mathrm{d}x=\frac{5}{34}e^{5x}\sin(3x)-\frac{3}{34}e^{5x}\cos(3x)+C$.

Answer:

$\frac{5}{34}e^{5x}\sin(3x)-\frac{3}{34}e^{5x}\cos(3x)+C$