# is there a nicer way to $\int e^{2x} \sin x\, dx$?

I have to solve this:

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

I managed to do it like this:

let $\space u_1 = \sin x \space$ and let $\space \dfrac{dv}{dx}_1 = e^{2x}$

$\therefore \dfrac{du}{dx}_1 = \cos x \space$ and $\space v_1 = \frac{1}{2}e^{2x}$

If I substitute these values into the general equation:

$\int u\dfrac{dv}{dx}dx = uv - \int v \dfrac{du}{dx}dx$

I get:

$\int e^{2x} \sin x dx = \frac{1}{2}e^{2x}\sin x - \frac{1}{2}\int e^{2x}\cos x\, dx$

Now I once again do integration by parts and say:

let $u_2 = \cos x$ and let $\dfrac{dv}{dx}_2 = e^{2x}$

$\therefore \dfrac{du}{dx}_2 = -\sin x$ and $v_2 = \frac{1}{2}e^{2x}$

If I once again substitute these values into the general equation I get:

$\int e^{2x}\sin x dx =\frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x - \frac{1}{4}\int e^{2x}\sin x dx$

$\therefore \int e^{2x}\sin x dx = \frac{4}{5}(\frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x) + C$

$\therefore \int e^{2x}\sin x\, dx = \frac{2}{5}e^{2x}\sin x - \frac{1}{5}e^{2x}\cos x + C$

I was just wondering whether there was a nicer and more efficient way to solve this?

Thank you :)

• Nope, this is pretty much the standard way to do it! (Unless you like complex numbers and write $\sin x = \frac1{2i}(e^{ix} - e^{-ix})$ - that's another way to do it - but few would call it nicer....) Jan 30 '14 at 16:19
• This is very well done ! There is another way using complex numbers. Do you want me to elaborate ? This is what Greg wrote ! Jan 30 '14 at 16:20
• Obviously, everyone is happy of what you did and propose the same alternate solution. Cheers. Jan 30 '14 at 16:22
• Essentially the same integral is asked about here: math.stackexchange.com/questions/136595/… Jan 30 '14 at 16:42

we have

$$\int e^{2x}\sin x \mathrm d x = (A\cos x + B\sin x)e^{2x}$$

Equation is the same at both ends of the derivative $x$

$$e^{2x}\sin x = 2e^{2x}(A\cos x +B \sin x) + (B\cos x -A\sin x)e^{2x}$$

Finishing $$e^{2x}\sin x =（2B-A)\sin x\,e^{2x} +(2A+B)\cos x\,e^{2x}$$

Compare coefficient，we have $$2A+B = 0\\ 2B-A = 1$$ Solutions have $$A =-\frac{1}{5} \\ B=\frac{2}{5}$$

• Ahh, this looks really good. Thanks. I'll have to look at it for a while to make sure I understand it. It's reminding me of complex roots when working with second order differentials but I'll have to look at it for a bit longer to understand how it's working :P Thanks again. Jan 30 '14 at 17:00
• If I may add a comment, this works because integration and derivation, seen as operators on the space of functions, both send the subspace of functions of that form considered into itself. Nice solution, +1. Jan 30 '14 at 20:30

\begin{align*} \\ \int e^{2x}\sin x dx &= \Im \int e^{2x}(\cos x + i\sin x) dx \\ &= \Im \int e^{(2 + i)x}dx \\ &= \Im \frac{e^{(2 + i)x}}{2+i} + C \\ &= \Im \frac15 e^{2x}(\cos x + i\sin x)(2-i) + C \\ &= \frac15 e^{2x}(2\sin x - \cos x) +C \end{align*}

• Thank you :) but I'm not too sure what the symbol in front of you integral sign is doing, is is just a constant like $\lambda$? I'm assuming it is what is allowing you to express $sinx$ as $cosx + isinx$ as I'm not too sure how this is done either. I know that $sin\theta = \frac{1}{2i}(e^{i\theta} - e^{-i\theta})$ have you manipulated this somehow? Thank you again :) Jan 30 '14 at 16:34
• These answers, this in particular, would be even nicer if you substitute $e^{2x}$ by $e^{nx}, \ n \neq 0$. Cheers! Jan 30 '14 at 16:37
• @Elise The symbol stands the "the imaginary part of" Jan 30 '14 at 16:38
• @JeffDror Right, okay thanks. I think I need to learn how to integrate complex numbers as this looks really cool :P It'll be a great way to double check my working in exams as well :D thanks everyone Jan 30 '14 at 16:44

A different method (though not really easier) is to use the fact that $\sin x$ can be expressed using exponents: $$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$$

So that your integral is: $$\int e^{2x} \sin x\ dx = \int \frac{e^{2x+ix} - e^{2x-ix}}{2i} dx= \frac{e^{2x+ix}}{4ix-2x} - \frac{e^{2x-ix}}{4ix+2x}+C$$ $$=\frac{-1}{10}(2i+1)e^{2x+ix} +\frac{1}{10} (2i-1)e^{2x-ix}+C$$ $$=\frac{2}{5}e^{2x}\sin x-\frac{1}{5}e^{2x}\cos x +C$$

• Thanks :D I can understand this up to $\int \dfrac{e^{2x+ix}-e^{2x-ix}}{2i}dx$ but unfortunately I don't know how to integrate imaginary numbers yet but I'm going to go learn now :D could you perhaps recommend a good book/website? Thanks again :) Jan 30 '14 at 16:47
• @Elise - there's nothing special going here, you can integrate imaginary numbers just like any other number, and at the end replace $i^2=-1$. Jan 30 '14 at 16:49
• Ahh, of course, thanks. I'll have another go :) Jan 30 '14 at 16:52