Question

Please help me finish this integral

Answer 1

yay integrals!

Answer 2

something more fun then algebra!

Answer 3

It is \[\int\limits_{}^{}e^x \sin(3x) dx\] so I made u = \[e^x\] du = e^x dv = sin(3x) v = -1/3 cos(3x)

Answer 4

then I got the following

Answer 5

\[\frac{ -1 }{ 3 }e^x \cos(3x) + \frac{ 1 }{ 3 }\int\limits_{}^{} \cos(3x) e^x dx\] then I did integration by parts again to get the following

Answer 6

what are the bounds? or is it an indefinite integral?

Answer 7

its indefinite

Answer 8

now I am stuck because it right back to the beginning

Answer 9

when you do integration by parts you end up with : \[u*v - \int\limits_{?}^{?}v du\]

Answer 10

\[\int\limits_{}^{}e^xsin(3x)dx = -\frac{ 1 }{ 3 }e^xcos(3x) + \frac{ 1 }{ 9 }e^x \sin(3x)-\frac{ 1 }{ 9 }\int\limits_{}^{}e^x \sin(3x)dx\]

Answer 11

try switching your u's and v's

Answer 12

I will end up with the same results

Answer 13

take e^x as dv
and take sin(3x) as u

Answer 14

@xapproachesinfinity

Answer 15

@ganeshie8

Answer 16

@johnnydicamillo
You're on the right track.
To finish,
Compare
\(\int\limits_{}^{}e^xsin(3x)dx = -\frac{ 1 }{ 3 }e^xcos(3x) + \frac{ 1 }{ 9 }e^x \sin(3x)-\frac{ 1 }{ 9 }\int\limits_{}^{}e^x \sin(3x)dx\)

with
\(x=-y+z - \dfrac{x}{9}\)
How would you solve for x in the last equation?

Answer 17

I have no idea = )

Answer 18

add the x/9

Answer 19

Can you solve the last equation for x in terms of y and z?

Answer 20

more hints will reveal the answer!

Answer 21

factor?

Answer 22

no never mind

Answer 23

I'm sorry I am not building the connection. You x/9 to the other side and add y to get z

Answer 24

Think of it this way. If \(I\) is the original integral, you have
\[I= -\frac{ 1 }{ 3 }e^xcos(3x) + \frac{ 1 }{ 9 }e^x \sin(3x)-\frac{ 1 }{ 9 }I\]
If you solve for \(I\), you will have computed the original integral. (Don't forget your constant.)