Arthur Pratt

2021-12-31

Is there a general formula for finding the primitive of
${e}^{ax}\mathrm{sin}\left(bx\right)$

Philip Williams

There is a pattern. Differentiating a function of the form ${e}^{ax}\mathrm{sin}\left(bx\right)$ yields a linear combination of a function of the same form, and a function ${e}^{ax}\mathrm{cos}\left(bx\right)$. The analogous property holds for functions ${e}^{ax}\mathrm{cos}\left(bx\right)$. So the primitive of ${e}^{ax}\mathrm{sin}\left(bx\right)$ will be a linear combination of ${e}^{ax}\mathrm{sin}\left(bx\right)$ and ${e}^{ax}\mathrm{cos}\left(bx\right)$ (plus a constant).
It remains to find the coefficients.
$\frac{d}{dx}\left({e}^{ax}\left(m\mathrm{sin}\left(bx\right)+n\mathrm{cos}\left(bx\right)\right)={e}^{ax}\left(a\left(m\mathrm{sin}\left(bx\right)+n\left(\mathrm{cos}\left(bx\right)\right)+\left(bm\mathrm{cos}\left(bx\right)-bn\mathrm{sin}\left(bx\right)\right)\right)$
$={e}^{ax}\left(\left(am-bn\right)\mathrm{sin}\left(bx\right)+\left(an+bm\right)\mathrm{cos}\left(bx\right)\right)$
Now solve the linear system
$am-bn=1$
$an+bm=0$

Barbara Meeker

Hint Integration by parts
$\int udv=uv-\int vdu$
Make substition
$u=\mathrm{sin}\left(bx\right)⇒du=b\mathrm{cos}\left(bx\right)dx$
and
$dv={e}^{ax}\mathrm{sin}\left(bx\right)dx=\frac{{e}^{ax}}{a}\mathrm{sin}\left(bx\right)-\frac{b}{a}\int {e}^{ax}\mathrm{cos}bxdx$
Then another integration by parts for $\int {e}^{ax}\mathrm{cos}bxdx$ . I think you can do the rest of it.

Vasquez

Try by parts twice (assuming $ab\ne 0$ to avoid trivialities):

and thus
$I:=\int {e}^{ax}\mathrm{sin}bxdx=-\frac{1}{b}{e}^{ax}\mathrm{cos}bx+\frac{a}{b}\int {e}^{ax}\mathrm{cos}bx$
$=-\frac{1}{b}{e}^{ax}\mathrm{cos}bx+\frac{a}{{b}^{2}}{e}^{ax}\mathrm{sin}bx-\frac{{a}^{2}}{{b}^{2}}\int {e}^{ax}\mathrm{sin}bxdx$
Well, now just past the last rightmost summand to the left side (above) and do a little algebra:
$\left(1+\frac{{a}^{2}}{{b}^{2}}\right)I=\frac{{e}^{ax}}{b}\left(\frac{1}{b}\mathrm{sin}bx-\mathrm{cos}bx\right)⇒I=...$

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