use integration by parts to evaluate the indefinite integral ∫e^6x sin(4x) dx.

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = \sin(4x)$ and $dv=e^{6x}dx$. Then $du = 4\cos(4x)dx$ and $v=\frac{1}{6}e^{6x}$. So, $\int e^{6x}\sin(4x)dx=\frac{1}{6}e^{6x}\sin(4x)-\int\frac{1}{6}e^{6x}\cdot4\cos(4x)dx=\frac{1}{6}e^{6x}\sin(4x)-\frac{2}{3}\int e^{6x}\cos(4x)dx$.

Step2: Apply integration - by - parts again

For $\int e^{6x}\cos(4x)dx$, let $u = \cos(4x)$ and $dv = e^{6x}dx$. Then $du=-4\sin(4x)dx$ and $v=\frac{1}{6}e^{6x}$. So, $\int e^{6x}\cos(4x)dx=\frac{1}{6}e^{6x}\cos(4x)+\int\frac{1}{6}e^{6x}\cdot4\sin(4x)dx=\frac{1}{6}e^{6x}\cos(4x)+\frac{2}{3}\int e^{6x}\sin(4x)dx$.

Step3: Substitute the second result into the first

Let $I=\int e^{6x}\sin(4x)dx$. We have $I=\frac{1}{6}e^{6x}\sin(4x)-\frac{2}{3}(\frac{1}{6}e^{6x}\cos(4x)+\frac{2}{3}I)$. Expand the right - hand side: $I=\frac{1}{6}e^{6x}\sin(4x)-\frac{1}{9}e^{6x}\cos(4x)-\frac{4}{9}I$.

Step4: Solve for $I$

Add $\frac{4}{9}I$ to both sides: $I+\frac{4}{9}I=\frac{1}{6}e^{6x}\sin(4x)-\frac{1}{9}e^{6x}\cos(4x)$. Combining like terms, $\frac{9I + 4I}{9}=\frac{1}{6}e^{6x}\sin(4x)-\frac{1}{9}e^{6x}\cos(4x)$, so $\frac{13}{9}I=\frac{1}{6}e^{6x}\sin(4x)-\frac{1}{9}e^{6x}\cos(4x)$. Multiply both sides by $\frac{9}{13}$: $I=\frac{3}{26}e^{6x}\sin(4x)-\frac{1}{13}e^{6x}\cos(4x)+C$.

Answer:

$\frac{3}{26}e^{6x}\sin(4x)-\frac{1}{13}e^{6x}\cos(4x)+C$