Question

solve this integration in details using integration by parts:(−e−x)cosxdx

Answer 1

\[\int\limits (-e ^{-x})cosx dx\]

Answer 2

\[-\int\limits (e ^{-x})cosx dx\]

Answer 3

\[\int\limits_{}^{}udv=uv-\int\limits_{}^{}vdu\]
u=cos(x)
\[dv=-e^{-x}\]

Answer 4

so:
\[\int\limits_{}^{}-e ^{-x}\cos(x)dx=e ^{-x} \cos(x)-\int\limits_{}^{}-e^{-x}\sin(x)\]

Answer 5

and
u=sin(x)
dv=-e^(-x)
so:
\[\int\limits_{}^{}-e^{-x}\cos(x)dx=e^{-x}\cos(x)-e^{-x}\sin(x)+\int\limits_{}^{}e^{-x}\cos(x)\]

Answer 6

subtract \[\int\limits_{}^{} e^{-x}\cos(x) \] from both sides:
\[-2\int\limits_{}^{}e^{-x}\cos(x)=e^{-x}\cos(x)-e^{-x}\sin(x)\]
divide each side by -2:
\[\int\limits\limits_{}^{}e^{-x}\cos(x)=(-1/2)(e^{-x}\cos(x)-e^{-x}\sin(x))+C\]