Question

Решите интеграл по частям

Answer 1

Ответ:

$\int e^{3x}cos 2x dx={9\over39}e^{3x}cos2x+\frac{18}{117}e^{3x}sin2x$

Объяснение:

Формула:

$\int {u} \, dv=uv-\int {v} \, du$

$\int e^{3x}cos 2x dx = \{u=cos2x, dv=e^{3x}\Rightarrow v= {e^{3x}\over3}, du=-2sin2x\}=\\={1\over3}e^{3x}cos2x+{2\over3}\int e^{3x}sin2x dx=\{u=sin2x, dv=e^{3x} \Rightarrow v={e^{3x}\over3},du=2cos2x\}={1\over3}e^{3x}cos2x+{2\over9}e^{3x}sin2x -{4\over9}\int e^{3x}cos 2x dx\\\\\int e^{3x}cos 2x dx={1\over3}e^{3x}cos2x+{2\over9}e^{3x}sin2x -{4\over9}\int e^{3x}cos 2x dx\\\\\frac{13}{9}\int e^{3x}cos 2x dx={1\over3}e^{3x}cos2x+{2\over9}e^{3x}sin2x$

$\int e^{3x}cos 2x dx={9\over39}e^{3x}cos2x+\frac{18}{117}e^{3x}sin2x$