The integral ($\int e^{ax} \sin(bx) \, dx$) is computed as:
\[
\int e^{ax} \sin(bx) \, dx= \frac{a e^{ax} \sin(bx)}{a^2 + b^2} – \frac{b e^{ax} \cos(bx)}{a^2 + b^2}+ C
\]
Proof of Integration
To solve the integral $\int e^{ax} \sin(bx) \, dx$, we can use the method of integration by parts, which is based on the formula:
$$
\int u \, dv = uv – \int v \, du
$$
Since the integral involves a product of two functions ($e^{ax}$ and $\sin(bx)$), and because direct integration isn’t straightforward, integration by parts is a suitable approach. In this case, we’ll apply the method twice, because the first application will bring us back to a similar integral form, allowing us to solve for the integral in terms of itself.
Let’s set:
- $u = e^{ax}$, which implies $du = a e^{ax} dx$
- $dv = \sin(bx) dx$, which implies $v = -\frac{1}{b} \cos(bx)$
Applying integration by parts, we get:
$$
\int e^{ax} \sin(bx) \, dx = -\frac{e^{ax}}{b} \cos(bx) – \int -\frac{a}{b} e^{ax} \cos(bx) \, dx
$$
For the second integral, we apply integration by parts again, this time with:
- $u = e^{ax}$ (again), resulting in $du = a e^{ax} dx$
- $dv = \cos(bx) dx$, resulting in $v = \frac{1}{b} \sin(bx)$
Thus, the integral becomes:
$$
-\frac{a}{b} \int e^{ax} \cos(bx) \, dx = -\frac{a}{b^2} e^{ax} \sin(bx) + \frac{a^2}{b^2} \int e^{ax} \sin(bx) \, dx
$$
Putting it all together, we have:
$$
\int e^{ax} \sin(bx) \, dx = -\frac{e^{ax}}{b} \cos(bx) + \frac{a}{b^2} e^{ax} \sin(bx) – \frac{a^2}{b^2} \int e^{ax} \sin(bx) \, dx
$$
Solving for the integral, we get:
$$
\left(1 + \frac{a^2}{b^2}\right) \int e^{ax} \sin(bx) \, dx = -\frac{e^{ax}}{b} \cos(bx) + \frac{a}{b^2} e^{ax} \sin(bx)
$$
Therefore, the integral is:
$$
\int e^{ax} \sin(bx) \, dx = \frac{-e^{ax} \cos(bx)}{b + \frac{a^2}{b}} + \frac{a e^{ax} \sin(bx)}{b^2 + a^2}
$$
The integral $\int e^{ax} \sin(bx) \, dx$ is simplified as
$$
\frac{a e^{ax} \sin(bx)}{a^2 + b^2} – \frac{b e^{ax} \cos(bx)}{a^2 + b^2}
$$
Solved Examples
Question 1
\[ \int e^x \sin(x) \, dx \]
Solution
Comparing to above formula , here a=1 and b =1, we get
\[ \int e^x \sin(x) \, dx = \frac{e^x (\sin(x) – \cos(x))}{2} + C \]
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