# Integrate $e^{ax}\sin(bx)?$

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

I've done this manually with $a=9$ and $b=4$ using Euler's formulas. But it takes a bit of time. Is there a pattern here?

• Mar 11, 2016 at 17:28

There is a pattern. Differentiating a function of the form $e^{ax}\sin (bx)$ yields a linear combination of a function of the same form, and a function $e^{ax}\cos (bx)$. The analogous property holds for functions $e^{ax}\cos (bx)$. So the primitive of $e^{ax}\sin (bx)$ will be a linear combination of $e^{ax}\sin (bx)$ and $e^{ax}\cos (bx)$ (plus a constant).

It remains to find the coefficients.

\begin{align} \frac{d}{dx}\left(e^{ax}(m\sin (bx) + n\cos (bx)\right) &= e^{ax}\left(a\bigl(m\sin(bx) + n\cos(bx)\bigr) + \bigl(bm\cos(bx) - bn\sin(bx)\bigr)\right)\\ &= e^{ax}\left((am - bn)\sin (bx) + (an+bm)\cos(bx)\right) \end{align}

Now solve the linear system

\begin{align} am - bn &= 1\\ an + bm &= 0. \end{align}

Notice that $$\sin(bx)=\mathrm{Im}(e^{ibx})$$

so we need just take the imaginary part of this antiderivative

$$\int e^{ax}e^{ibx}dx=\frac{1}{a+ib}e^{(a+ib)x}+C$$

Hint Integration by parts

$$\int u \ dv =uv-\int v \ du$$

Make substition $$u=\sin(bx)\ \Rightarrow \ du=b\cos (bx) \ dx$$ and $$\ dv=e^{ax} \ dx \Rightarrow v= \frac{e^{ax}}{a}$$ So $$\int e^{ax} \sin(bx) \ dx=\frac{e^{ax}}{a}\sin(bx)-\frac ba\int e^{ax}\cos bx \ dx$$

Then another integration by parts for $\int e^{ax}\cos bx \ dx$ . I think you can do the rest of it.

• The OP said he already did this. This answer helps him with nothing. Oct 26, 2013 at 13:31
– TMM
Oct 26, 2013 at 13:33
• @TMM Apparently he did not think of integration by parts yet? To me it is obvious he did so. Oct 26, 2013 at 13:35
• "I've done this manually with a=9 and b=4 using Euler's formulas." - I guess it's not user1772257 who did not read the question.
– TMM
Oct 26, 2013 at 13:36
• @TMM You're right. Oct 26, 2013 at 13:38

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

$$u=e^{ax}\;\;,\;\;u'=ae^{ax}\\ v'=\sin bx\;\;,\;\;v=-\frac1b\cos bx$$

and thus

$$I:=\int e^{ax}\sin bx\,dx=-\frac1be^{ax}\cos bx+\frac ab\int e^{ax}\cos bx\,dx=$$

$$=-\frac1be^{ax}\cos bx+\frac a{b^2}e^{ax}\sin bx-\frac{a^2}{b^2}\int e^{ax}\sin bx\,dx$$

Well, now just past the last rightmost summand to the left side (above) and do a little algebra:

$$\left(1+\frac{a^2}{b^2}\right)I=\frac{e^{ax}}b\left(\frac 1b\sin bx-\cos bx\right)\implies I=\ldots$$

• Shouldn't it be $\left(1+\frac{a^2}{b^2}\right)I=\frac{e^{ax}}b\left(\frac{\color{red}{a}}b\sin bx-\cos bx\right)$? May 14, 2021 at 19:19

Write $\sin bx = \Im e^{ibx}$, so that

$$e^{ax} \sin bx = \Im e^{ax}e^{ibx}=e^{(a+ib)x}.$$

Integrate this as a regular exponential and recover the imaginary part:

$$\int e^{ax} \sin (bx) dx = \int \Im e^{ax}e^{ibx}dx= \Im \int e^{(a+ib)x}dx.$$

$e^{ax}(a\sin(bx)-b\cos(bx))/(a^2+b^2)$

• Welcome to math.stackexchange. Since this question already has many other answers, it would be good to add more details to your answer, explaining how it is different from the others.
– mrf
Oct 12, 2015 at 11:42