yuki
  • yuki
find the integral
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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yuki
  • yuki
\[\int\limits_{0}^{1} e^x \sin(x) dx\]
anonymous
  • anonymous
\[(1/2)[e(\sin1-\cos1)+1\]
yuki
  • yuki
how did you get that ?

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yuki
  • yuki
I was thinking of integration by parts
anonymous
  • anonymous
use by parts
yuki
  • yuki
letting e^xdx = dv and u = sin(x)
yuki
  • yuki
that makes e^x = v and du = sin(x)dx
yuki
  • yuki
thus making uv -vdu = e^x cos(x) - [int] e^x cos(x) dx
yuki
  • yuki
du = cos(x) dx, my bad
yuki
  • yuki
so now I need to figure out what \[\int\limits e^x \cos(x) dx\]
yuki
  • yuki
is equal to
yuki
  • yuki
so using same stuff, I would get -e^x sin(x) - [int] e^x sin(x) dx
yuki
  • yuki
since I get the same thing on both sides, I can add them on both sides, right?
anonymous
  • anonymous
u can use dis..the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (-beta cos(beta x)+alpha sin(beta x)))/(alpha^2+beta^2):
anonymous
  • anonymous
my idea is that sinx= Im(e^ix) so the integral is Im( e^(i+1)x)
anonymous
  • anonymous
that is Im(e^i+1)x/(i+1)
anonymous
  • anonymous
now put back 1 and 0 (e^(i+1)-1)/i+1
anonymous
  • anonymous
what do you think guys?
anonymous
  • anonymous
I am stuck here.. what is e^i+1?
anonymous
  • anonymous
yuki..u r right u can do by using by parts..see \[\int\limits_{0}^{x}e^xsin(x)dx=[e^xsin(x)]-\int\limits_{0}^{x}e^xcos(x)dx=e^xsin(x)-[e^xcos(x)+\int\limits_{0}^{x}e^xsin(x)dx]\] \[2\int\limits_{0}^{x}e^xsin(x)dx=e^xsin(x)-e^xcos(x)\] Now Put x=1 in the above..:)
anonymous
  • anonymous
nice!

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