Question

int e^2x(cos3x)dx please help :)

Answer 1

:( lol

Answer 2

u=cos3x?

Answer 3

but i feel like there might be some trig identity at play here

Answer 4

ah yeah batman

Answer 5

At first glance integration by parts comes to mind.

Answer 6

ok so if i set u=cos3x

Answer 7

Have you tried? Let u = cos3x

Answer 8

then du is what 1/3(sin3x)?

Answer 9

integration by part is what i think of when i see this

Answer 10

You're taking the derivative

Answer 11

Not integrating cos3x

Answer 12

so 3sin3x?

Answer 13

Try again :)

Answer 14

sin(3x)*(3x)' right?

Answer 15

or is it -sin

Answer 16

Hehe nice

Answer 17

lmao GRR TRIGGGG

Answer 18

v = ?

Answer 19

v=e2x?

Answer 20

good going^_^

Answer 21

Ye homeslice dawg my mang dis is how we do 1t

Answer 22

omg batman just got real that's what im talking bout

Answer 23

think of... LIPETs. (log, integral, polynomial, exponential, trig) when doing partials

(delayed post)

Answer 24

LIPETs hit me back with some memories^_^!!!

Answer 25

:)

Answer 26

ok wow so i end up with \[e ^{2x}\cos(3x)-\int\limits_{}^{}-3e^{2x}(sinx)dx\]

Answer 27

so then i take out the -3

Answer 28

sin3x not sinx

Answer 29

oh yeah sorry. thats what i have on paper, just made the eq bad over here

Answer 30

this a hick of a intergral by part you will repeat the integration several times

Answer 31

so i have now

Answer 32

\[e^{2x}\cos(3x)+3e^{2x}\sin(3x)-9\int\limits_{}^{}e^{2x}\cos(3x)\]

Answer 33

Usually this problem you continue till something repeats and thereafter you need to cancel some term to finish it off

Answer 34

can i just add the 9inte^2xcos3x to the other side

Answer 35

very good observation my friend!
that's how it work you throw that term to the other side

Answer 36

sweet. just throwing me off cause that 9 lol.

Answer 37

I'm not sure you did everything correctly! but you steps are descent and that's how to throw off this intergration

Answer 38

this one turned to be less steps than i anticipated! As I said some of them
you need to carry on tell you recognize repeatition

Answer 39

Then you just label the repetition with a letter.

Answer 40

label the repetition with a letter?

Answer 41

Yeah like a big Letter A
say A=........-9A
A is your starting integral
or could be something else

Answer 42

you are doing great you know!
continue....

Answer 43

i ended with (1/10)(e^2x)(cos3x+3sin3x)+c
but wolfram says it should be (1/13)(e^2x)(2cos3x+3sin3x)+c

Answer 44

and i can't figure out what i did wrong

Answer 45

i started with ine^2xcos3x
u=cos3x du=-3sin3x dv=e^2xdx v=e^2x

Answer 46

so then i got e^2xcos3x+3inte^2x(sin3x)dx
u=sin3x du=3cos3x dv=e^2xdx v=e^2x

Answer 47

so e^2xcos3x+3[e^2xsin3x-int3e^2xcosxdx]

Answer 48

just go back to your calculations and see if there is something wrong
and beside i don't really trust wolfram always
but for this one i this wolfram is right
the problem is that wolfram goes deep in integrating