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Finding $$\int e^{2x} \sin 4x \, dx$$

I think I should be doing integration by parts...

If I let $u=e^{2x} \Rightarrow du = 2e^{2x}$,
$dv = \sin{4x} \Rightarrow v = -\frac{1}{4} \cos{4x}$
$\int{ e^{2x} \sin{4x}} dx = e^{2x}(-\frac{1}{4}\cos 4x) - \color{red}{\int (-\frac{1}{4} \cos 4x) 2e^{2x} \, dx}$ Highlighted in red, I seem to integrating an exponential times trig expression again ... doesn't seem like te right way to go

So I let $u=\sin{4x} \Rightarrow du = 4\cos 4x \, dx$
$dv = e^{2x} dx \Rightarrow v = \frac{1}{2} e^{2x}$
$\sin{4x}(\frac{1}{2} e^{2x}) - \int (\frac{1}{2} e^{2x})(4\cos4x) \, dx$
Again, its an exponential times a trig function?

Am I using the wrong substitution?

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Yes, after one integration by parts you are getting an exponential times trig expression again, but that doesn't mean it is not the right way to go! Just apply by parts again and enjoy. –  Shitikanth Apr 25 '12 at 3:18
    
This is similar to the integral of $e^x \cos x$, which wikipedia has worked-out. I'd suggest taking a look at that and applying the principles (which "countinghaus" mentioned, more or less) –  The Chaz 2.0 Apr 25 '12 at 3:30
2  
I wonder how many times this same question has been asked here. In a way, you could say every instructor who sets this particular problem intends to provoke this particular question. (Unless the instructor is very naive, which certainly does happen.) (Anyway, I've posted an answer below.) –  Michael Hardy Apr 25 '12 at 4:04
    
And, most probably, your calculus textbook has a worked-out example or two where you integrate by parts twice and then solve. –  GEdgar Apr 26 '12 at 17:29

5 Answers 5

up vote 10 down vote accepted

I wonder how many times this same question has been asked here, and how many times it is asked every semester in every second-semester calculus course?

If you integrate by parts a second time, you get the same integral you started with, and the naive reaction is that that means you're getting nowhere. But in truth it means you're almost done. You've got $$ \int \text{something} = \text{something} - \text{something}\cdot \int [\text{same integral}]. $$ So you add the same thing to both sides of the equation and get $$ \int \text{something} + \text{something}\cdot\int [\text{same integral}] = \text{something} + C $$ Then you write $$ (1+\text{something}) \cdot\int\cdots\cdots = \text{something} + C $$ $$ \int\cdots\cdots = \frac{\text{something}}{1+\text{something}} + \text{constant}. $$

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Take your red integral and integrate it by parts again, making sure to let $u$ be the exponential function again. You will get that

The integral you care about = stuff + the integral you care about*(other stuff)

and therefore

the integral you care about = stuff/(1 - other stuff)

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In general, if you want to find $$ \int e^{ax}\cdot \sin{bx}\cdot dx$$ you can argue as follows:

Note that for any $\alpha$ or $\beta$, you have

$$\eqalign{ & \frac{d}{{dx}}\left( {{e^{\alpha x}}\sin \beta x} \right) = \alpha {e^{\alpha x}}\sin \beta x + \beta {e^{\alpha x}}\cos \beta x \cr & \frac{d}{{dx}}\left( {{e^{\alpha x}}\cos \beta x} \right) = \alpha {e^{\alpha x}}\cos \beta x - \beta {e^{\alpha x}}\sin \beta x \cr} $$

so that any integral of the form

$$ \int e^{\alpha x}\cdot \sin{\beta x}\cdot dx$$

is a linear combination of the former functions. Let's then find $c_1$ and $c_2$ such that

$$\frac{d}{{dx}}\left( {{c_1}{e^{\alpha x}}\sin \beta x + {c_2}{e^{\alpha x}}\cos \beta x} \right) = {e^{\alpha x}}\sin \beta x$$

$${c_1}\alpha {e^{\alpha x}}\sin \beta x + {c_1}\beta {e^{\alpha x}}\cos \beta x + {c_2}\alpha {e^{\alpha x}}\cos \beta x - {c_2}\beta {e^{\alpha x}}\sin \beta x = {e^{\alpha x}}\sin \beta x$$

This means we need

$$\eqalign{ & {c_1}\alpha - {c_2}\beta = 1 \cr & {c_1}\beta + {c_2}\alpha = 0 \cr} $$

This will yield with little work

$$\eqalign{ & {c_1} = \frac{\alpha }{{{\alpha ^2} + {\beta ^2}}} \cr & {c_2} = - \frac{\beta }{{{\alpha ^2} + {\beta ^2}}} \cr} $$

which means that, in general:

$$\int {{e^{\alpha x}}} \cdot\sin \beta x\cdot dx = {e^{\alpha x}}\frac{{\alpha \sin \beta x - \beta \cos \beta x}}{{{\alpha ^2} + {\beta ^2}}} + C$$

Analogously, you will get that

$$\int {{e^{\alpha x}}} \cdot\cos \beta x\cdot dx = {e^{\alpha x}}\frac{{\alpha \cos \beta x + \beta \sin \beta x}}{{{\alpha ^2} + {\beta ^2}}} + C$$

Hope this helps!

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Integrating with derivatives and linear algebra, I like it. (+1) –  user26872 Apr 26 '12 at 23:10

I would jump into complex values, and noting that $\sin 4x = \Im e^{4 i x}$ (I wrote \Im there, wanting to get the imaginary part, and got the $\Im$),

$$\int e^{2x} \sin 4x \, dx = \Im \int e^{2x+4ix} \, dx = \Im \int e^{x(2+4i)} \, dx = \Im \frac{e^{x(2+4i)}}{2+4i}. $$

Since $\frac{1}{2+4i} = \frac{2-4i}{20} = \frac{1-2i}{10}$ and $e^{x(2+4i)} = e^{2x}(\cos 4x + i \sin 4x)$, I get, disregarding the $\frac{e^{2x}}{10}$ for a moment,

$\Im (1-2i)(\cos 4x + i \sin 4x) = \Im ((\cos 4x + 2 \sin 4x) + i(-2\cos 4x + \sin 4x)) $ $ = -2\cos 4x + \sin 4x $

so my final result is $\frac{e^{2x}(-2\cos 4x + \sin 4x)}{10}$.

As a check the derivative of $e^{2x}(-2\cos 4x + \sin 4x)$ (disregarding the 1/10 for now) is

$$ \begin{align} & 2e^{2x}(-2\cos 4x + \sin 4x) + e^{2x}(8\sin 4x + 4\cos 4x) \\[4pt] = {} & e^{2x}(-4\cos 4x + 2\sin 4x + 8\sin 4x + 4\cos 4x) = e^{2x} (10 \sin 4x). \end{align} $$

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1  
I would not jump into complex values, in case anyone is curious. –  The Chaz 2.0 Apr 25 '12 at 5:04
4  
When I do, I wear a parachute and a portable air bag. This minimizes the landing impact. –  marty cohen Apr 26 '12 at 5:05

To find $\int e^{2x}\sin4x\,dx,$ we can let $u=e^{2x}$ and $dv=\sin4x\,dx,$ which implies $du=2e^{2x}\,dx$ and $v=-\frac{1}{4}\cos4x.$ Therefore, $$\begin{align}\int e^{2x}\sin4x\,dx &= uv-\int v\,du\\ &= e^{2x}\left(-\frac{1}{4}\cos4x\right)-\int-\frac{1}{4}\cos4x\left(2e^{2x}\right)\,dx\\ &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{2}\int e^{2x}\cos4x\,dx.\end{align}$$ Apply another integration for $\int e^{2x}\cos4x\,dx,$ this time letting $u=e^{2x}$ and $dv=\cos4x\,dx,$ so that $du=2e^{2x}\,dx$ and $v=\frac{1}{4}\sin4x.$ Hence, $$\begin{align}\int e^{2x}\cos4x\,dx &= uv-\int v\,du\\ &= e^{2x}\left(\frac{1}{4}\sin4x\right)-\int\frac{1}{4}\sin4x\left(2e^{2x}\right)\,dx\\ &= \frac{1}{4}e^{2x}\sin4x-\frac{1}{2}\int e^{2x}\sin4x\,dx.\end{align}$$

Combining both formulas, we get $$\begin{align}\int e^{2x}\sin4x\,dx &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{2}\left(\frac{1}{4}e^{2x}\sin4x-\frac{1}{2}\int e^{2x}\sin4x\,dx\right)\\ &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{8}e^{2x}\sin4x-\frac{1}{4}\int e^{2x}\sin4x\,dx.\end{align}$$ So, $$\frac{5}{4}\int e^{2x}\sin4x\,dx=-\frac{1}{4}e^{2x}\cos4x+\frac{1}{8}e^{2x}\sin4x.$$ Therefore, multiplying by $\frac{4}{5}$ and adding in the constant of integration, we see that $$\int e^{2x}\sin4x\,dx=-\frac{1}{5}e^{2x}\cos4x+\frac{1}{10}e^{2x}\sin4x+C,$$ which we can rewrite as $$\int e^{2x}\sin4x\,dx=\frac{1}{10}e^{2x}\left(-2\cos4x+\sin4x\right)+C.$$

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Hi, welcome to Math.SE. Our site supports the use of MathJax for mathematical formulas, and the use of MarkDown for formatting text. I would highly recommend your editing your post to make use of both to improve on the clarity of exposition. –  Willie Wong Jan 17 at 16:02
    
Welcome to math.SE: I have tried to improve the readability of your post by introducing Tex. It is possible that I unintentionally changed the meaning of your post. Please proofread the question to ensure this has not happened. It's also worth noting that this question is nearly two years old, and has an accepted answer, so is probably too late to "help" the original poster. –  Cameron Buie Jan 17 at 16:28

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