19. $\\int e^{x}\\cos 2x dx$

Explanation:

Step1: Use integration by parts

Let (u = \cos2x), (dv=e^{x}dx). Then (du=- 2\sin2x dx), (v = e^{x}). By integration by parts formula (\int u;dv=uv-\int v;du), we have (\int e^{x}\cos2x dx=e^{x}\cos2x + 2\int e^{x}\sin2x dx).

Step2: Use integration by parts again

For (\int e^{x}\sin2x dx), let (u = \sin2x), (dv=e^{x}dx). Then (du = 2\cos2x dx), (v=e^{x}). By integration by parts formula (\int e^{x}\sin2x dx=e^{x}\sin2x-2\int e^{x}\cos2x dx).

Step3: Substitute back

Substitute (\int e^{x}\sin2x dx) into the first - integration - by - parts result: (\int e^{x}\cos2x dx=e^{x}\cos2x + 2\left(e^{x}\sin2x-2\int e^{x}\cos2x dx\right)) (\int e^{x}\cos2x dx=e^{x}\cos2x + 2e^{x}\sin2x-4\int e^{x}\cos2x dx)

Step4: Solve for (\int e^{x}\cos2x dx)

Add (4\int e^{x}\cos2x dx) to both sides: (5\int e^{x}\cos2x dx=e^{x}\cos2x + 2e^{x}\sin2x) (\int e^{x}\cos2x dx=\frac{1}{5}e^{x}(\cos2x + 2\sin2x)+C)

Answer:

(\frac{1}{5}e^{x}(\cos2x + 2\sin2x)+C)