question evaluate the integral below. ∫e^4x cos(5x)dx be sure to place the argument of any trigonometric or…

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = \cos(5x)$ and $dv=e^{4x}dx$. Then $du=- 5\sin(5x)dx$ and $v=\frac{1}{4}e^{4x}$. So, $\int e^{4x}\cos(5x)dx=\frac{1}{4}e^{4x}\cos(5x)+\frac{5}{4}\int e^{4x}\sin(5x)dx$.

Step2: Apply integration - by - parts again on $\int e^{4x}\sin(5x)dx$

Let $u = \sin(5x)$ and $dv = e^{4x}dx$. Then $du = 5\cos(5x)dx$ and $v=\frac{1}{4}e^{4x}$. So, $\int e^{4x}\sin(5x)dx=\frac{1}{4}e^{4x}\sin(5x)-\frac{5}{4}\int e^{4x}\cos(5x)dx$.

Step3: Substitute the result of Step2 into Step1

$\int e^{4x}\cos(5x)dx=\frac{1}{4}e^{4x}\cos(5x)+\frac{5}{4}\left(\frac{1}{4}e^{4x}\sin(5x)-\frac{5}{4}\int e^{4x}\cos(5x)dx\right)$. $\int e^{4x}\cos(5x)dx=\frac{1}{4}e^{4x}\cos(5x)+\frac{5}{16}e^{4x}\sin(5x)-\frac{25}{16}\int e^{4x}\cos(5x)dx$.

Step4: Solve for $\int e^{4x}\cos(5x)dx$

Add $\frac{25}{16}\int e^{4x}\cos(5x)dx$ to both sides: $\left(1 + \frac{25}{16}\right)\int e^{4x}\cos(5x)dx=\frac{1}{4}e^{4x}\cos(5x)+\frac{5}{16}e^{4x}\sin(5x)$. $\frac{41}{16}\int e^{4x}\cos(5x)dx=\frac{1}{4}e^{4x}\cos(5x)+\frac{5}{16}e^{4x}\sin(5x)$. $\int e^{4x}\cos(5x)dx=\frac{4}{41}e^{4x}\cos(5x)+\frac{5}{41}e^{4x}\sin(5x)+C$.

Answer:

$\frac{4}{41}e^{4x}\cos(5x)+\frac{5}{41}e^{4x}\sin(5x)+C$