# How to integrate a three products

I tried to integrate $x e^x \sin x$, using integration by parts, and setting $dv/dx = e^x \sin x$. Even though I got really close, I kept getting it wrong. Can someone please solve it with working out? Thanks in advance.

***EDIT**** I have found the answer, all thanks to those who contributed :) I don't know if I could of connected the answer here but I just posted it below, thank you all again!

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this aint that straight forward. –  Lost1 Jan 13 at 11:07
Im not sure if you can read my writing but i think i got pretty close to the answer "wolframalpha", @Lost1. fluorine-silver.blogspot.com.au/2014/01/math-so-damn-close.html –  Samir Chahine Jan 13 at 11:11
i dont understand your v and v'. this question is such a pain in the arse if you don't use the hint Nigel posted, but needs complex numbers. I wanted to post a solution using elementary integration and i got stuck... –  Lost1 Jan 13 at 11:14
–  lab bhattacharjee Jan 13 at 11:15
There is no such thing as a "three product". The function $x\mapsto x\mathrm e^x\sin x$ is a function, dot. You could write it as well $2\sqrt x\left(\mathrm e^{x/2}\right)^2\sqrt x\sin(x/2)\cos(x/2)$ and it would be a six function... –  Tom-Tom Jan 13 at 16:53

If you don't want to try with complex calculus, you might use this:

Let's define $g(x) = e^x \sin{x}$, so we have: $J = \int x g(x) \, dx$.

Then, if you use chain rule, you will have:

$$J = x\, \int g(x) \, dx - \int \! \int g(x) \, dx^2,$$

where, using again the chain rule:

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

Thus:

$$\int \! \int g(x) dx^2 = \int \frac{e^x}{2} (\sin{x} - \cos{x}) \, dx$$

where the first integral has already been computed. Using again the chain rule for the cosine integral, it finally yields:

$$J = \frac{x}{2} e^x (\sin{x} - \cos{x} ) + \frac{e^x}{2} \cos{x}$$

Do not forget the integration constant! Cheers.

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yes this is nice +1 –  Lost1 Jan 13 at 11:26
Sorry, I made a mistake on the transcription. –  Dmoreno Jan 13 at 11:32

Let me elaborate on Nigel's hint, and btw he meant $e^{ix} = i \sin x +\cos x$. There is no $\pi$.

then the integral you want is J. define the integral $I = \int x \cos x e^x \text{d}x$.

Then $I + iJ = \int x e^{ix}e^x \text{d}x = \int xe^{(i+1)x}\text{d}x$

note $i$ is square root of minus one, so it i just a constant, you integrate this by part and seperate out real and imaginary part. The imaginary part is what you want.

At first I wanted to post this, but I guessed this is not what OP wanted.

however, without this, this integral is a pain in the backside...

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@Ruslan lol.... –  Lost1 Jan 13 at 11:20
Guess I have to learn complex numbers, thank you, if you do find out a way to solve this without it please do post it :) –  Samir Chahine Jan 13 at 11:20
@SamirChahine there is, i am sure you can crack it if you give it another 3 hours, using integration by parts... –  Lost1 Jan 13 at 11:21
I will do it, and when I do I will post it, I shall never give up. @Lost1, may I ask how long it would take to learn complex numbers? And do I need to know any other type of math before I dive in –  Samir Chahine Jan 13 at 11:21
@SamirChahine take $du/dx = x, v = e^x \sin x$ as your first step, and work from there –  Lost1 Jan 13 at 11:22

Hint: $$\sin(\theta)= Im (e^{i \theta})$$ Then use integration by parts.

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I have absolutely no idea what that is, a hint for this hint please? –  Samir Chahine Jan 13 at 11:12
Hint for the hint: use Euler formula. –  Ruslan Jan 13 at 11:15
You're absolutely right. Thanks and fixed it. –  Nigel Overmars Jan 13 at 11:21

Hint with complex numbers: $$\sin(\theta) = \frac{ e^{i\theta} - e^{-i \theta} }{2i}$$

Without complex numbers: Let $f = x$, $g' = e^x \sin x$.

First we calculate $g = \int g' dx$ by integration by parts: $$\begin{array}{c} I = \int {{e^{ax}}\sin (bx)dx} = \left[ {\begin{array}{*{20}{c}} {u = \sin (bx)}&{v' = {e^{ax}}}\\ {u' = b\cos (bx)}&{v = {e^{ax}}/a} \end{array}} \right]\mathop = \limits^{{\mathop{\rm int}} } \frac{{{e^{ax}}\sin (bx)}}{a} - \int {b\cos (bx)\frac{{{e^{ax}}}}{a}dx} \\ = \frac{{{e^{ax}}\sin (bx)}}{a} - \frac{b}{a}\int {\cos (bx){e^{ax}}dx} = \\ = \left[ {\begin{array}{*{20}{c}} {f = \cos (bx)}&{g' = {e^{ax}}}\\ {f' = - b\sin (bx)}&{g = {e^{ax}}/a} \end{array}} \right] = \frac{{{e^{ax}}\sin (bx)}}{a} - \frac{b}{a}\left( {\frac{{\cos (bx){e^{ax}}}}{a} - \frac{{ - b}}{a}\int {\sin (bx){e^{ax}}dx} } \right) = \\ = \frac{{{e^{ax}}\sin (bx)}}{a} - \frac{{b\cos (bx){e^{ax}}}}{{{a^2}}} - \frac{{{b^2}}}{{{a^2}}}\int {\sin (bx){e^{ax}}dx} = \frac{{{e^{ax}}\sin (bx)}}{a} - \frac{{b\cos (bx){e^{ax}}}}{{{a^2}}} - \frac{{{b^2}}}{{{a^2}}}I \end{array}$$ Thus $$I = \frac{{{e^{ax}}\sin (bx)}}{a} - \frac{{b\cos (bx){e^{ax}}}}{{{a^2}}} - \frac{{{b^2}}}{{{a^2}}}I$$ Solving for $I$ we get $$I = \frac{{{e^{ax}}\left( {a\sin (bx) - b\cos (bx)} \right)}}{{{a^2} + {b^2}}} + C$$ so $$g(x) = \frac{{{e^{x}}\left( {\sin (x) -\cos (x)} \right)}}{{2}} \ .$$

Thus, to solve the big integral we do again integration by parts with $f=x$: $$\int f g' = fg - \int f' g = x \frac{{{e^{x}}\left( {\sin (x) -\cos (x)} \right)}}{{2}} - \int \left( \frac{{{e^{x}}\left( {\sin (x) -\cos (x)} \right)}}{{2}} \right) dx$$ where the last integral can be calculated as above.

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Nice double differentiation of $\sin$ :) –  Ruslan Jan 13 at 11:23

One general idea with products of three functions is to use the product rule in the form $$(u v w)' = u' v w + u v' w + uv w'$$ and the get partial integration in the form $$\int u' v w = uvw - \int u v' w - \int uv w'$$ and then the solution of your problem is straightforward but tedious.

After two applications of above rule (with $u=e^x$) and some reorganization you find
$$2 \int x e^x \sin x \, dx = xe^x \sin x - x e^x \cos x -\int e^x \sin x \, dx + \int e^x \cos x \, dx$$ and the rest is easy.

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

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

$$u = x \\ u'= 1\\\\v'=e^x\sin(x) \\v= \frac{1}{2}e^x(\sin x -\cos x) + C \\ I = \frac{xe^x}{2}(\sin x-\cos x )-\frac{1}{2} \int e^x(\sin x-\cos x) + C \\ \text{Let }Z^- = \sin x -\cos x \\ \text{and}\\ \text{Let }Z^+ = \sin x+\cos x \\ \frac {xe^x}{2}(Z^-) - \frac{1}{2}\left(\frac{1}{2}e^x(Z^- - Z^+)\right) +C\\ \frac {xe^x}{2}(\sin x-\cos x) - \frac{1}{2}\left(\frac{1}{2}e^x(-2\cos x )\right)+C\\ \frac {xe^x}{2}(\sin x-\cos x) + \frac{1}{2}e^x(\cos x) + C\\ \int xe^x\cos x\ dx = \frac{1}{2}e^x(x\sin x-x\cos x+\cos x) + C$$ Well. I told you guys I would do it, thank you all for your help, turn out I've been stuck on this bastard for 3 days not two, time flies when you're having fun! ;) - I entered most of the steps, you guys should see the gaps if there are any, and i'm sorry my MathJax isn't perfect, this is my first time using it for such a big equation, thank you all again! Next stop, five function ;) (I checked on wolframalpha.com, got the same answer!)

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