Question

integrate e^-x cos2x dx ????

Answer 1

try by parts :)

Answer 2

Have you tried a method?

Answer 3

It can do by twice times to get the solution ^^

Answer 4

because i am asking, It has no analytical solution i think

Answer 5

Do I need to show the solution ???

Answer 6

ok you are right

Answer 7

first I define u = cos(2x) and dv = e^-x dx\[\int\limits_{}^{}udv=\int\limits_{}^{}e ^{-x}\cos2xdx\]du=2cos(2x)dx and v = -e^-x using the formula \[\int\limits_{}^{}udv=uv-\int\limits_{}^{}vdu=-e ^{-x}\cos2x-\int\limits_{}^{}\left( -e ^{-x}\right)\left( -\sin2x \right)dx\]do by parts on second term again with u' = sin(2x) and dv' = e^-x ,thus du' = 2cos(2x), v' = -e^-x\[\int\limits_{}^{}e ^{-x}\sin2x=-e ^{-x}\sin2x-\int\limits_{}^{}\left( -e ^{-x} \right)\left( 2cos2x \right)dx\]substitute into the second equation \[\int\limits_{}^{}e ^{-x}\cos2xdx=-e ^{-x}\cos2x-\left[ -e ^{-x}\sin2x+\int\limits_{}^{}2e ^{-x}\cos2xdx \right]\]you'll see that on the both side have int(e^-x*cos2xdx) so rearrange to obtain\[3\int\limits_{}^{}e ^{x}\cos2xdx=-e ^{-x}\left( \cos2x-\sin2x \right)\]
and the solution is\[\int\limits_{}^{}e ^{x}\cos2xdx=\frac{ -1 }{3 }e ^{x}\left( \cos2x-\sin2x \right)\]

Answer 8

on the last twoe equation exp(x) must be exp(-x) I jus type it wrong

Answer 9

hi how do you get step by step solution. can share the link?

Answer 10

I just do it myself ><, has no any program for it :P