Fórum de Matemática
DÚVIDAS? Nós respondemos!

\(\int e^3^Xcos(7x)dx\)


ajudem pf não consigo resolver

\(\\\\ \int e^3^xcos(7x)dx \\\\ u=cos (7x) \rightarrow du=-7*sen(7x) \\ dv=e^{3x}\rightarrow v=\frac{e^{3x}}{3} \\\\ \int u dv=u*v-\int v du \\\\ \int e^3^xcos(7x)dx =\frac{e^{3x}*cos(7x)}{3}+\int \frac{e^{3x}}{3}*7sen(7x)dx\)

agora repare que temos novamente uma integral por partes vamos resolver separadamente:

\(\\\\\\ \int e^{3x}sen(7x)dx \\\\ u=sen(7x)\rightarrow du=7*cos(7x) \\ dv=e^{3x}\rightarrow v=\frac{e^{3x}}{3}dx \\\\\\ \int e^{3x}sen(7x)=sen(7x)*\frac{e^{3x}}{3}-\int \frac{e^{3x}}{3}*7cos(7x)dx\)


vamos substituir esse resultado na primeira integral:

\(\\\\\\ \int e^3^xcos(7x)dx =\frac{e^{3x}*cos(7x)}{3}+\frac{7}{3}*\int e^{3x}sen(7x)dx \\\\\\ \int e^3^xcos(7x)dx =\frac{e^{3x}*cos(7x)}{3}+\frac{7}{3}(sen(7x)*\frac{e^{3x}}{3}-\frac {7 } {3 }\int e^{3x}cos(7x)dx) \\\\\\ \int e^3^xcos(7x)dx =\frac{e^{3x}*cos(7x)}{3}+\frac{7*sen(7x)*e^{3x}}{9}-\frac {49 } {9 }\int e^{3x}cos(7x)dx \\\\\\ \int e^3^xcos(7x)dx\frac {49 } {9 }+\int e^{3x}cos(7x)dx =\frac{e^{3x}*cos(7x)}{3}+\frac{7*sen(7x)*e^{3x}}{9} \\\\\\ \frac{ 58}{9}\int e^{3x}cos(7x)dx =\frac{3*e^{3x}*cos(7x)+7*sen(7x)*e^{3x}}{9} \\\\\\ \int e^{3x}cos(7x)dx =(3*e^{3x}*cos(7x)+7*sen(7x)*e^{3x})*58\)

att .