Question

integral indefinida e^3x cos 4x dx

Answer 1

$\int e^{3x}*cos(4x) dx$

integrando por partes

$\int UdV = U*V - \int VdU\\\\U=cos(4x)\\dU=-4sen(4x)dx\\\\dV=e^{3x}dx\\V= \frac{e^{3x}}{3}\\\\ \boxed{\boxed{ \int e^{3x}cos(4x) dx = \frac{e^{3x}}{3}cos(4x) + \frac{4}{3} \int e^{3x}sen(4x)dx}} \to \text{ eqauacao 1}$

resolvendo a outra integral que surgiu

$\int e^{3x}sen(4x)dx\\\\ U=sen(4x)\\dU=4cos(4x)dx\\dV=e^{3x}\\V= \frac{e^{3x}}{3} \\\\ \boxed{\boxed{\int e^{3x}sen(4x)dx= \frac{e^{3x}}{3} sen(4x)- \frac{4}{3} \int e^{3x}cos(4x)dx}}$

substituindo o valor da integral na equação 1

$\int e^{3x}cos(4x) dx = \frac{e^{3x}}{3}cos(4x) + \frac{4}{3}\left(\frac{e^{3x}}{3} sen(4x)- \frac{4}{3} \int e^{3x}cos(4x)dx\right)\\\\ \int e^{3x}cos(4x) dx = \frac{e^{3x}}{3}cos(4x) + \frac{4e^{3x}}{9} sen(4x)- \frac{16}{9} \int e^{3x}cos(4x)dx\\\\\ \int e^{3x}cos(4x) dx + \frac{16}{9} \int e^{3x}cos(4x) dx = \frac{e^{3x}}{3}cos(4x) + \frac{4e^{3x}}{9} sen(4x)\\\\ \frac{25}{9} \int e^{3x}cos(4x) dx= \frac{3e^{3x}cos(4x)+4e^{3x}sen(4x)}{9}$

$\boxed{\boxed{ \int e^{3x}cos(4x) dx= \frac{e^{3x}}{25}*\left[3cos(4x)+4sen(4x)\right] }}$