Question

∫sin(y)/y dy

Answer 1

and your question? How?

Answer 2

integration by parts

Answer 3

yeah
how do i integrate
\[\int\limits_{}^{}{siny \over y} dy\]

Answer 4

make one your u

Answer 5

and the other your dv

Answer 6

will that work? what is a good choice for u,v'

Answer 7

i'd use 1/y as v'

Answer 8

sin(y) as u

Answer 9

thanks
, il try that

Answer 10

\[\int\limits_{}^{}{siny \over y}= \ln y siny| -\int\limits_{}^{}\ln(y)\cos(y)dy\]

Answer 11

this looks more complicated to me
have i have done this correctly
whats the next step

Answer 12

sometimes with integration by parts you have to do it until you get the same integral term on both sides

Answer 13

do it one more time

Answer 14

you'll end up with a negative ∫siny/y on the right

Answer 15

move it to the left, divide by 2

Answer 16

and you'll have your answer. It's the only way you'll get a 'simple' answer. (this is called the sine integral, actually)

Answer 17

http://www.wolframalpha.com/input/?i=%5C%5BIntegral%5DSin%5By%5D%2Fy+%5C%5BDifferentialD%5Dy

Answer 18

yes, but no. That is a /form/ of the answer, but not generally recognisable as such (as in, it would probably /not/ count for homework). Hence integration by parts.

Answer 19

i'll let you get an answer and then i gotta go

Answer 20

\[\int_{}^{}{\sin y \over y}= \ln y \sin y| -\int_{}^{} \ln y\cos y dy=\ln y \sin y| -\ln y \sin y| +\int_{}^{} {\sin y \over y} dy = \int_{}^{} {\sin y \over y} dy\]

Answer 21

\[=\int_{}^{} {\sin y \over y} dy\]

Answer 22

that's not right.

Answer 23

too complicated. ;)

Answer 24

well i gotago right now

Answer 25

k cool bye!

Answer 26

=Si(x)
???