Krdytkyu

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!

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.

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.

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...

Hint, using complex numbers:
$$ sin(theta) = frac{ e^{itheta} - e^{-i theta} }{2i} $$

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

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