Question

$\int\ {}e^2^x~senx \, dx$

Answer 1

Boa noite!

Solução!

Integral por partes cíclica!

$\displaystyle \int udv=uv- \displaystyle \int vdu\\\\\\\\\ \displaystyle \int e^{2x} senxdx\\\\\\\\\ u=sen(x)dx~~~~~dv=e^{2x} \\\\\\\\\ du=cos(x)dx~~~~~v= \dfrac{e^{2x} }{2}\\\\\\ \displaystyle \int e^{2x} senxdx= \dfrac{sen(x).e^{2x}}{2}-\displaystyle \int\dfrac{e^{2x} }{2}cos(x)dx\\\\\\\ \displaystyle \int e^{2x} senxdx= \dfrac{sen(x).e^{2x}}{2}- \dfrac{1}{2} \displaystyle \int e^{2x} cos(x)dx$

Vamos fazer a mesma operação para a integral do lado direito!

$\displaystyle \int e^{2x} cos(x)dx\\\\\\\ u=cos(x)~~~~dv=e^{2x} \\\\\\\ du=-sen(x)dx~~~~v= \dfrac{e^{2x} }{2}\\\\\\ \int udv=uv- \displaystyle \int vdu\\\\\\\\\\ \dfrac{cos(x).e^{2x} }{2}- \dfrac{1}{2} \displaystyle \int e^{2x}[-sen(x)dx]\\\\\\ \dfrac{cos(x).e^{2x} }{2}+ \dfrac{1}{2} \displaystyle \int e^{2x}sen(x)dx$


Vamos agora juntar as duas partes!

$\displaystyle \int e^{2x} senxdx=\dfrac{sen(x).e^{2x}}{2}-\dfrac{cos(x).e^{2x} }{4}-\dfrac{1}{4} \displaystyle \int e^{2x}sen(x)dx\\\\\\\\\ \displaystyle \int e^{2x} senxdx+\dfrac{1}{4} \displaystyle \int e^{2x}sen(x)dx=\dfrac{sen(x).e^{2x}}{2}-\dfrac{cos(x).e^{2x} }{4}$


$\dfrac{5}{4} \displaystyle \int e^{2x}sen(x)dx= \dfrac{sen(x).e^{2x}}{2}-\dfrac{cos(x).e^{2x} }{4}\\\\\\\\ \displaystyle \int e^{2x}sen(x)dx= \frac{\dfrac{sen(x).e^{2x}}{2}-\dfrac{cos(x).e^{2x} }{4}}{ \dfrac{5}{4} }\\\\\\\\\ \displaystyle \int e^{2x}sen(x)dx= \dfrac{sen(x).e^{2x}}{2}-\dfrac{cos(x).e^{2x} }{4}. \frac{4}{5} \\\\\\\\\ \displaystyle \int e^{2x}sen(x)dx= \dfrac{4sen(x).e^{2x}}{10}-\dfrac{4cos(x).e^{2x} }{20}$


$\displaystyle \int e^{2x}sen(x)dx= \dfrac{2sen(x).e^{2x}}{5}-\dfrac{1cos(x).e^{2x} }{5} \\\\\\\\\\ \displaystyle \int e^{2x}sen(x)dx= \frac{1}{5} . \left (2sen(x).e^{2x}-cos(x).e^{2x}\right ) +c \\\\\\\\\\ \boxed{Resposta:~~\displaystyle \int e^{2x}sen(x)dx= \frac{1}{5} . \left (2sen(x).e^{2x}-cos(x).e^{2x}\right ) +c}$

Boa noite!
Bons estudos!

Answer 2

$\sf \displaystyle \int \:e^{2x}sin \left(x\right)dx\\\\\\=-e^{2x}\cos \left(x\right)-\int \:-2e^{2x}\cos \left(x\right)dx\\\\\\=-e^{2x}\cos \left(x\right)-\left(-2\cdot \int \:e^{2x}\cos \left(x\right)dx\right)\\\\\\=-e^{2x}\cos \left(x\right)-\left(-2\left(e^{2x}\sin \left(x\right)-\int \:e^{2x}\cdot \:2\sin \left(x\right)dx\right)\right)\\\\\\=-e^{2x}\cos \left(x\right)-\left(-2\left(e^{2x}\sin \left(x\right)-2\cdot \int \:e^{2x}\sin \left(x\right)dx\right)\right)$

$\sf \displaystyle \int \:e^{2x}\sin \left(x\right)dx=-e^{2x}\cos \left(x\right)-\left(-2\left(e^{2x}\sin \left(x\right)-2\cdot \int \:e^{2x}\sin \left(x\right)dx\right)\right)~~\\\\\\=-\frac{e^{2x}\cos \left(x\right)}{5}+\frac{2e^{2x}\sin \left(x\right)}{5}\\\\\\\to \boxed{\sf =-\frac{e^{2x}\cos \left(x\right)}{5}+\frac{2e^{2x}\sin \left(x\right)}{5}+C}$