How to integrate a three products

Question

I tried to integrate $x e^x \sin x$, using integration by parts, and setting $\frac{\, \mathrm dv}{\, \mathrm dx} = e^x \sin x$. 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!

Answer 1

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

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

$$ u = x \\ u'= 1\\\\v'=e^x\sin(x) \\v= \frac{1}{2}e^x(\sin x -\cos x) + C \\ I = \frac{xe^x}{2}(\sin x-\cos x )-\frac{1}{2} \int e^x(\sin x-\cos x) + C \\ \text{Let }Z^- = \sin x -\cos x \\ \text{and}\\ \text{Let }Z^+ = \sin x+\cos x \\ \frac {xe^x}{2}(Z^-) - \frac{1}{2}\left(\frac{1}{2}e^x(Z^- - Z^+)\right) +C\\ \frac {xe^x}{2}(\sin x-\cos x) - \frac{1}{2}\left(\frac{1}{2}e^x(-2\cos x )\right)+C\\ \frac {xe^x}{2}(\sin x-\cos x) + \frac{1}{2}e^x(\cos x) + C\\ \int xe^x\cos x\ dx = \frac{1}{2}e^x(x\sin x-x\cos x+\cos x) + C $$ 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!)

Answer 2

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

Answer 3

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...

Answer 4

Hint, using 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.

Answer 5

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:

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

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:

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

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

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

Do not forget the integration constant! Cheers.

Answer 6

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.