Question

find the integral

Answer 1

\[\int\limits_{0}^{1} e^x \sin(x) dx\]

Answer 2

\[(1/2)[e(\sin1-\cos1)+1\]

Answer 3

how did you get that ?

Answer 4

I was thinking of integration by parts

Answer 5

use by parts

Answer 6

letting e^xdx = dv and u = sin(x)

Answer 7

that makes e^x = v and du = sin(x)dx

Answer 8

thus making uv -vdu = e^x cos(x) - [int] e^x cos(x) dx

Answer 9

du = cos(x) dx, my bad

Answer 10

so now I need to figure out what
\[\int\limits e^x \cos(x) dx\]

Answer 11

is equal to

Answer 12

so using same stuff, I would get

-e^x sin(x) - [int] e^x sin(x) dx

Answer 13

since I get the same thing on both sides, I can add them on both sides, right?

Answer 14

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):

Answer 15

my idea is that sinx= Im(e^ix)
so the integral is Im( e^(i+1)x)

Answer 16

that is Im(e^i+1)x/(i+1)

Answer 17

now put back 1 and 0
(e^(i+1)-1)/i+1

Answer 18

what do you think guys?

Answer 19

I am stuck here.. what is e^i+1?

Answer 20

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..:)

Answer 21

nice!