## University Calculus: Early Transcendentals (3rd Edition)

$$\int e^{2x}\cos3xdx=\frac{2}{13}e^{2x}\cos3x+\frac{3}{13}e^{2x}\sin3x+C$$
$$A=\int e^{2x}\cos3xdx$$ Set $u=\cos3x$ and $dv=e^{2x}dx$ Then we have $du=-3\sin 3xdx$ and $v=\frac{1}{2}e^{2x}$ Using the formula $\int udv= uv-\int vdu$: $$A=\frac{1}{2}e^{2x}\cos3x-\int\frac{1}{2}e^{2x}\times(-3)\sin3xdx$$ $$A=\frac{1}{2}e^{2x}\cos3x+\frac{3}{2}\int e^{2x}\sin3xdx$$ Set $u=\sin3x$ and $dv=e^{2x}dx$ Then we have $du=3\cos 3xdx$ and $v=\frac{1}{2}e^{2x}$ Using the formula $\int udv= uv-\int vdu$: $$A=\frac{1}{2}e^{2x}\cos3x+\frac{3}{2}\Big(\frac{1}{2}e^{2x}\sin3x-\frac{3}{2}\int e^{2x}\cos3xdx\Big)$$ We notice that above, $\int e^{2x}\cos3xdx$ is exactly the given integral to solve, meaning it equals $A$. Therefore, $$A=\frac{1}{2}e^{2x}\cos3x+\frac{3}{2}\Big(\frac{1}{2}e^{2x}\sin3x-\frac{3}{2}A\Big)$$ $$A=\frac{1}{2}e^{2x}\cos3x+\frac{3}{4}e^{2x}\sin3x-\frac{9}{4}A$$ $$\frac{13}{4}A=\frac{1}{2}e^{2x}\cos3x+\frac{3}{4}e^{2x}\sin3x+C$$ $$A=\frac{2}{13}e^{2x}\cos3x+\frac{3}{13}e^{2x}\sin3x+C$$