Question

Calculate the integral e^(2x)sin(2x)dx

Answer 1

Is sin(2x) also raised to the power?

Answer 2

is this one of those parts in a circle problems?

Answer 3

nope just e

nope its just studying for my calc 2 exam and im not sure if i can just solve this any way...such as integration by parts?

Answer 4

@satellite73 yes :3

Answer 5

parts and then i think you go around in circles i.e. get back what you started with, then divide by 2

Answer 6

try integrating by parts and see what happens
i loathe integration with a passion, mostly because it is a complete show off waste of time

Answer 7

how can i solve this integral quickly and efficiently

Answer 8

keep in mind I'm studying for exam:D

Answer 9

actually i am not sure if it makes any difference what you choose for \(u\) and \(dv\) you can try \(u=e^x, du = e^xdx, dv = \sin(x), v=-\cos(x)\) and get
\[-\cos(x)e^x+\int e^x\cos(x)dx\]

Answer 10

no there is no real quick way
then parts again

Answer 11

correct...and then solve for the integral of e^(x)cos(x)

Answer 12

can i just put up the questions and u can check my answers?

Answer 13

that means i would have to do them and i would make a mistake
quickest bet is to use wolfram to check

Answer 14

okay thanks

Answer 15

http://www.wolframalpha.com/input/?i=e^%282x%29sin%282x%29

Answer 16

but you get the idea right? when you use parts again, you are going to get the original integral back, then you solve for it

Answer 17

yeah

Answer 18

\[\int e^x\sin(x)=-\cos(x)e^x+\int e^x\cos(x)dx=\]\[-\cos(x)e^x+e^x\sin(x)-\int e^x\sin(x)dx\]

Answer 19

i dropped the \(2x\) so i guess you have to be more careful with the constants

Answer 20

where did u get the e^(x)sin(x) on the left side?

Answer 21

it was the original question

Answer 22

i mean the original integral

Answer 23

why would u add the original equation?

Answer 24

you end up with \[\int e^x\sin(x)dx=-\cos(x)e^x+e^x\sin(x)-\int e^x\sin(x)dx\] and so
\[2\int e^x\sin(x)=-\cos(x)e^x+e^x\sin(x)\] then divide by 2

Answer 25

i.e. when you integrate by parts twice, the second time you get the minus the original integral back, then you solve for it

Answer 26

okay

Answer 27

more or less clear?

Answer 28

nope but its fine ill ask someone tomorrow...

Answer 29

we can do it slow if you like, this one is not so hard

Answer 30

im beyond frustrated with this work....they gave us a study sheet but no answer key....

no its fine thank you though

Answer 31

aw :c

Answer 32

ok yw
but just to try
\[\int e^x\sin(x)dx=-\cos(x)e^x+e^x\sin(x)-\int e^x\sin(x)dx\] is like
\[Y=-\cos(x)e^x+e^x\sin(x)-Y\] solve for \(Y\)

Answer 33

add \(Y\) to both sides, divide both sides by 2
that will get it

Answer 34

im guna work it out and see if i get anything close to the right answer...

Answer 35

I dont understand how youre explaining it im sorry

Answer 36

if I have the integral of 2cos(2x)*e^(2x) can I take out both 2s and put them outside the integral?

@satellite73

Answer 37

Yes,\[\Large \int\limits 2\cos(2x) e^{2x}\;dx \quad=\quad 2\int\limits \cos(2x)e^{2x}\;dx\]
I'm not sure what you mean by `both` 2s. :o

Answer 38

you answered my question...thank you:)