Question

Integral de e^2x sen(x)dx

Answer 1

$\large\boxed{\begin{array}{l}\displaystyle\sf\int e^{2x}\cdot sen(x)\,dx\\\sf u= sen(x)\implies du=cos(x)dx\\\sf dv=e^{2x}\,dx\implies v=\dfrac{1}{2}e^{2x}\\\displaystyle\sf\int e^{2x}\cdot sen(x)\,dx= sen(x)\cdot\dfrac{1}{2}e^{2x}-\int\dfrac{1}{2}e^{2x}\cdot cos(x)\,dx\\\displaystyle\sf\int e^{2x}\cdot sen(x)\,dx=\dfrac{1}{2}e^{2x}\cdot sen(x)-\dfrac{1}{2}\int e^{2x}\cdot cos(x)\,dx\end{array}}}$

$\large\boxed{\begin{array}{l}\displaystyle\tt\int e^{2x}\cdot cos(x)\,dx\\\tt u_1= cos(x)\,dx\implies du_1=-sen(x)\,dx\\\tt dv_1= e^{2x}\,dx\implies v_1=\dfrac{1}{2}e^{2x}\\\displaystyle\tt\int e^{2x}\cdot cos(x)\,dx= cos(x)\cdot\dfrac{1}{2}e^{2x}-\int\dfrac{1}{2}e^{2x}\cdot -sen(x)\,dx\\\displaystyle\tt\int e^{2x}\cdot cos(x)\,dx=cos(x)\cdot\dfrac{1}{2}e^{2x}+\dfrac{1}{2}\int e^{2x}\cdot sen(x)\,dx\end{array}}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int e^{2x}\cdot sen(x)\,dx= \dfrac{1}{2}e^{2x}\cdot sen(x)\,-\dfrac{1}{2}\bigg[\dfrac{1}{2}e^{2x}\cdot cos(x)+\dfrac{1}{2}\int e^{2x}\cdot sen(x)\,dx\bigg]\\\displaystyle\sf\int e^{2x}\cdot sen(x)\,dx=\dfrac{1}{2}e^{2x}\cdot sen(x)-\dfrac{1}{4}e^{2x}\cdot cos(x)-\dfrac{1}{4}\int e^{2x}\cdot sen(x)\,dx\\\displaystyle\sf\int e^{2x}\cdot sen(x)\,dx+\dfrac{1}{4}\int e^{2x}\cdot sen(x)\,dx=\dfrac{1}{2}e^{2x} sen(x)-\dfrac{1}{4}e^{2x}cos(x)\end{array}}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\dfrac{5}{4}\int e^{2x}sen(x)\,dx=\dfrac{1}{2}e^{2x}sen(x)-\dfrac{1}{4}e^{2x}cos(x)\\\displaystyle\sf\int e^{2x}\cdot sen(x)\,dx=\dfrac{4}{5}\bigg[\dfrac{1}{2}e^{2x} sen(x)-\dfrac{1}{4}e^{2x}cos(x)\bigg]+k\\\displaystyle\sf\int e^{2x}sen(x)\,dx=\dfrac{2}{5}e^{2x}sen(x)-\dfrac{1}{5}e^{2x}cos(x)+k\\\displaystyle\sf\int e^{2x}sen(x)\,dx=\dfrac{1}{5}e^{2x}(2sen(x)-cos(x))+k\end{array}}}$