question evaluate the integral below. ∫e^2x cos(4x) dx be sure to place the argument of any trigonometric or…

Explanation:

Step1: Apply integration - by - parts

Let $u = \cos(4x)$ and $dv=e^{2x}dx$. Then $du=-4\sin(4x)dx$ and $v=\frac{1}{2}e^{2x}$. By the integration - by - parts formula $\int u;dv = uv-\int v;du$, we have: $\int e^{2x}\cos(4x)dx=\frac{1}{2}e^{2x}\cos(4x)+ 2\int e^{2x}\sin(4x)dx$.

Step2: Apply integration - by - parts again on $\int e^{2x}\sin(4x)dx$

Let $u = \sin(4x)$ and $dv = e^{2x}dx$. Then $du = 4\cos(4x)dx$ and $v=\frac{1}{2}e^{2x}$. $\int e^{2x}\sin(4x)dx=\frac{1}{2}e^{2x}\sin(4x)-2\int e^{2x}\cos(4x)dx$.

Step3: Substitute the result of Step2 into Step1

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

Step4: Solve for $\int e^{2x}\cos(4x)dx$

Add $4\int e^{2x}\cos(4x)dx$ to both sides: $5\int e^{2x}\cos(4x)dx=\frac{1}{2}e^{2x}\cos(4x)+e^{2x}\sin(4x)$. $\int e^{2x}\cos(4x)dx=\frac{1}{10}e^{2x}\cos(4x)+\frac{1}{5}e^{2x}\sin(4x)+C$.

Answer:

$\frac{1}{10}e^{2x}\cos(4x)+\frac{1}{5}e^{2x}\sin(4x)+C$