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

Question

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?

Answer 1

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

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

Answer 2

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

so we need just take the imaginary part of this antiderivative

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

Answer 3

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.

Answer 4

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:

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

Answer 5

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:

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

Answer 6

There are couple of ways to do this such as IBP(Integration by parts) but the method I will be using is by using complex numbers:

Using euler's identity which is:

$$e^{ix} = \cos(x) + i \sin(x)$$

We can say that:

$$e^{-ix} = \cos(-x) + i \sin(-x)$$

Cosine is an even function and sine is an odd function so this means that:

$$e^{-ix} = \cos(x) - i \sin(x)$$

Since we are looking for what sin(x) is we multiply the equation by -1:

$$e^{ix} = \cos(x) + i \sin(x)$$ $$-e^{-ix} = -\cos(x) + i \sin(x)$$

The cosines cancel out so your left with:

$$e^{ix}-e^{-ix} = {2i}\sin(x)$$

Divide both sides by 2i and we can say that:

$$\frac{e^{ix}-e^{-ix} } {2i}=\sin(x)$$

If we go back to the integral and substitute this result instead of sin(bx) we end up with:

$$\int e^{ax}(\frac{e^{ibx}-e^{-ibx} } {2i}) dx $$

The $\frac{1} {2i}$ is just a constant so we can factor that out and we can use properties of indices to rewrite the integral in the following way:

$$\frac{1} {2i}\int e^{ax+ibx}-e^{ax-ibx} dx $$

If we use the sum-difference rule we can seperate the integrand into two different integrals then if we factor out the x:

$$\frac{1} {2i}\int e^{x(a+ib)}-\frac{1} {2i}\int e^{x(a-ib)} dx $$

We can integrate the two integrals seperately with integration by inspection and finally end up with:

$$\frac{e^{(a+bi)x}} {2i(a+bi)}\ - \frac{e^{(a-bi)x}} {2i(a-bi)}\ + c $$

Answer 7

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