How to integrate $\int e^{-x}\sin(3x)\;dx$?

Question

I want to integrate following by the method of integration by parts

$$\frac{\cos(3x)}{e^{x}}$$

when I try to solve it by integration by parts it always leads to something like as mentioned below and it still have integration sign around it.

$$\frac{\sin(3x)}{e^{x}}$$

Please help I am really confused.

Any reason you are dividing by $e^{-x}$ instead of multiplying by $e^x$? Also: You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. — Zev Chonoles, Jun 08 '13 at 10:36
just bring $e^{-x}$ to numerator. You will be able to do it with integration by parts. You have to do it twice. — Maesumi, Jun 08 '13 at 10:36
http://math.stackexchange.com/questions/408515/liate-ilate-rule/408554#408554 try this(check my answer) — iostream007, Jun 08 '13 at 10:38
Is the present form perhaps the question that you intended to ask? — Lord_Farin, Jun 08 '13 at 10:45
http://www.proofwiki.org/wiki/Integral_of_Product_of_Exponential_with_Sine_or_Cosine — lab bhattacharjee, Jun 08 '13 at 10:47
Always try to apply integration by parts when you encounter such functions made up of terms which exhibit periodicity with respect to repeated integration/differentiation. — Thomas, Jun 08 '13 at 14:24

score 6 · Answer 1 · edited Jun 08 '13 at 12:15

6

we have $\int \sin(3x)e^{-x}dx$ so we take $f=\sin(3x),g'=e^{-x}$ getting $$=-\sin(3x)e^{-x}+3\int \cos(3x)e^{-x}dx$$

and integrating by parts again ($f=\cos(3x),g'=e^{-x}$) gives $$\int \cos(3x)e^{-x}dx=-\cos(3x)e^{-x}-3\int \sin(3x)e^{-x}dx$$ or generally we can conclude $$\int \sin(3x)e^{-x}=-\sin(3x)e^{-x}-3\cos(3x)e^{-x}-9\int \sin(3x)e^{-x} dx$$ and then that's clear that (if I didn't have any calculation mistake):$$\int \sin(3x)e^{-x}dx=\frac {-3\cos(3x)e^{-x}-\sin(3x)e^{-x}}{10}+C$$

edited Jun 08 '13 at 12:15

robjohn

345,667

answered Jun 08 '13 at 10:47

There were some presentational issues; the integrand is actually $\sin(3x)e^x$. Also, \sin (resp. \cos) in math mode gives $\sin$ ($\cos$), which is better than $sin$ ($cos$). – Lord_Farin Jun 08 '13 at 10:49
3

@DanielMargolis: Your edit was correct. Unfortunately, it was mistakenly rejected. I have made your changes, but there is no way to get you your points for the suggested edit. – robjohn Jun 08 '13 at 12:19

score 3 · Answer 2 · answered Jun 08 '13 at 10:43

3

$$\int e^{3ix}e^{-x}dx=\int e^{(3i-1)x}dx=\frac{1}{3i-1}e^{(3i-1)x}+C=-\frac{1+3i}{10}e^{-x}(\cos(3x)+i\sin(3x))+C$$ Now expand and take the imaginary part.

answered Jun 08 '13 at 10:43

Now... why do we take the imaginary part? – apnorton Jun 08 '13 at 11:42
@anorton Because you are integrating $\sin$ and $\sin$ is the imaginary part of this when you expand it. – Kaish Jun 08 '13 at 11:59
1

@anorton If it helps looking at it this way, $\int \Im(f(x))dx=\Im(\int f(x)dx)$. – Meow Jun 08 '13 at 12:21

Kaish · Answer 3 · 2013-06-08T11:49:12.583

The way I write integration by parts is by first choosing one of the functions to be $d$, which is the one we differentiate, and the other function to be $i$, which is the one we integrate. Then, we integrate by parts by doing the following: $d \times \int i - \int \{\int i \cdot $ differential of $d \}$. Which, I think is the same as the formula, but I don't really know how to use the formula properly because I was taught to think of it in this way. Now for your integral, we start of by saying let

$$ I =\int e^{-x} \sin(3x)dx$$

If we select $e^{-x}$ to be the one we integrate and so we will differentiate $\sin(3x)$, as that is easier. So now, if we apply my by parts algorithm to calculate $ I$, we start off by holding $\sin(3x)$, multiplying it by $\int e^{-x}dx = -e^{-x}$, and then subtracting the integral of $-\int e^{-x} \cdot 3 \cos(3x)$, so we get

$$I = \sin(3x) \cdot - e^{-x} - \int \left( - \int e^{-x} \cdot 3 \cdot \cos(3x) \right)$$ $$ = - \sin(3x) e^{-x} - 3 \int \left(e^{-x} \cos(3x)\right)dx.$$

Now, once again, we apply the by parts algorithm to the new bit involving $\cos$. Again, we can choose to integrate $e^{-x}$ and differentiate $\cos(3x)$ and so we get

$$ I = - \sin(3x)e^{-x} - 3 \left[ \cos(3x) \cdot -e^{-x} - \int \left( - \int e^{-x} \cdot - 3\sin(3x) \right) \right]$$

tidying this up gives us

$$I = - \sin(3x)e^{-x} - 3 \left[-\cos(3x)e^{-x} + 3 \underbrace{ \int \left(e^{-x} \sin(3x) \right) }_\text{=I} \right].$$

From here, we see that the final bit in the integral is our original integral, i.e $I$, so we can substitute this in and simplify to get

$$I = - \sin(3x)e^{-x} -3 (- \cos(3x)e^{-x} + 3I)$$ $$I = - \sin(3x)e^{-x} + 3 \cos(3x)e^{-x} - 9I.$$

We can now re-arrange and solve for $I$, which is the answer to our original integral:

$$10I = - \sin(3x)e^{-x} + 3\cos(3x) e^{-x}$$ $$I = \frac{e^{-x} \left( 3 \cos(3x) - \sin(3x) \right)}{10}$$

which is your final answer.

Daniel Margolis · Answer 4 · 2013-06-08T12:04:41.927

0

$$\int e^{3ix}e^{-x}dx=\int e^{(3i-1)x}dx=\frac{1}{3i-1}e^{(3i-1)x}+C=-\frac{1+3i}{10}e^{-x}(\cos(3x)+i\sin(3x))+C=\frac{-\sin(3x)-3\cos(3x)}{10}e^{-x}+C$$

All but the first guy is wrong. The 3 cos is negative.

edited Jun 08 '13 at 12:04

answered Jun 08 '13 at 11:57

Daniel Margolis

351

you are right. that was a mistake. – Jun 08 '13 at 13:05

How to integrate $\int e^{-x}\sin(3x)\;dx$?

4 Answers4