How to integrate a three products

Solution 1:

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.

Solution 2:

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.

Solution 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...

Solution 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.