Question

Integrate by parts int_{}^{}e ^{2x}sin(3x)dx

Answer 1

\[\int\limits_{}^{}e ^{2x}Sin(3x)dx\]

Answer 2

\[\int e^{2x}sin(3x)dx\]\[=\frac{1}{2}\int sin(3x)d(e^{2x})\]Then, integration by parts. Can you do it?

Answer 3

You need to do integration by parts twice.

Answer 4

\[\frac{ -1 }{ 3 }e ^{2x}\cos(3x)+\frac{ 2 }{ 3 }\int\limits_{}^{}e ^{2x}\cos(3x)\]

Answer 5

Hmm...\[\int udv = uv-\int vdu\]

Answer 6

I've been using\[u=e ^{2x}, du= 2e ^{2x}\]\[v=-\frac{ 1 }{ 3 }\cos(3x), dv=\sin(3x)\] did I do the first part right?

Answer 7

Hmm.. I don't recommend this way :|

Answer 8

Should I have used cos(3x) for u instead?

Answer 9

Do you follow this formula \(\int udv = uv-\int vdu\)?

Answer 10

Yes

Answer 11

Let v = (1/2) e^(2x)
u = sin(3x)

Answer 12

From the question, you have \(\int uv'dx =\int e ^{2x}sin(3x)dx\)
So, Let u = sin(3x) => u' =...
v' = e^(2x) => v =...

Answer 13

\[u'=3\cos(3x)\]\[v=\frac{ 1 }{ 2 }e ^{2x}\]\[\frac{ 1 }{ 2 }e ^{2x}\sin(3x)-\int\limits_{}^{}\frac{ 3 }{ 2 }e ^{2x}\cos(3x)\]

Answer 14

Yup and do integration by parts again..

Answer 15

\[\frac{ 1 }{ 2 }e ^{2x}\sin(3x)-\frac{ 3 }{ 2 }(\frac{ 1 }{ 2 }e ^{2x}\cos(3x)+\frac{ 3 }{ 2 }\int\limits_{}^{}e ^{2x}\sin(3x))\]

Answer 16

Yup! So, you have
\[\int e^{2x}sin(3x)dx = \frac{ 1 }{ 2 }e ^{2x}\sin(3x)-\frac{ 3 }{ 2 }(\frac{ 1 }{ 2 }e ^{2x}\cos(3x)+\frac{ 3 }{ 2 }\int\limits_{}^{}e ^{2x}\sin(3x))\]Add \(\frac{9}{4}\int e^{2x}sin(3x)dx\) to both sides. What will you get?

Answer 17

\[\frac{ 1 }{ 2 }e ^{2x}\sin(3x)-\frac{ 3 }{ 4 }e ^{2x}\cos(3x)=\int\limits_{}^{}e ^{2x}\sin(3x)dx +\frac{ 9 }{ 4 }\int\limits_{}^{}e ^{2x}\sin(3x)dx\]

Answer 18

\[=\frac{ 13 }{ 4 }\int\limits_{}^{}e ^{2x}\sin(3x)dx\]

Answer 19

Divide both sides by 13/4, then add a constant. You should be able to get the answer

Answer 20

Ahh I see now, I just kept integrating by parts and got confused when it never ended

Answer 21

And now you did it :)

Answer 22

Thank you