Question

Help with this integral

Answer 1

When I try using parts, it becomes endless...

Answer 2

When you integral the second time, it comes back to the original one, stop there!

Answer 3

how do I solve this

Answer 4

How do choose u and dv?

Answer 5

So that the integral simplifies

Answer 6

maybe I can make dv = dx and u = e^2x sinx?

Answer 7

So, you totally don't know about integral by Part :(

Answer 8

du = 2e^2x
v = -cosx

Answer 9

-cosx(e^2x) - (integral (-cosx)(2e^2x))

Answer 10

u = cosx, dv = e^(2x) ?

Answer 11

what have you showed me? all you did was say u = e^2x and dv = sinx and told me to use parts

Answer 12

So how do you do it?

Answer 13

so now I just substitute u for e^2x?

Answer 14

integral (ucosx)?

Answer 15

u = e^2x , dv = cosx

Answer 16

du = 2e^(2x)
v = sinx

Answer 17

no it never ends...

Answer 18

Plug in, you'll see!
I've told you that stop at the second time, have I?

Answer 19

ok

e^(2x)(sinx) - 2 integral((sinx)(e^(2x)))

Answer 20

yes

Answer 21

i don't know

Answer 22

why did you use I ?

Answer 23

ok what do I do with I

Answer 24

(-cosx)(e^(2x)) + 2(e^(2x)sinx - 2I)

Answer 25

5 I = -e^(2x) cosx + 2e^(2x) sinx

Answer 26

I = (-e^(2x) cosx + 2e^(2x) sinx) / 5

Answer 27

Again, simplify:
I = [ e^(2x)/ 5 ] [ 2sinx - cosx ]

Answer 28

how did you get that?

Answer 29

factor e^(2x) out, that's all :0

Answer 30

oh ok I see. now what

Answer 31

"NOW WHAT" , that's DONE !!!

Answer 32

wow this is interesting. what is the name of this technique?

Answer 33

and why does it work

Answer 34

nevermind i figured it out. Because you set it equal to the beginning so you can say everything equals I, then solve for I. Wow thanks!