Example 6. Find $\int e^{\alpha x} \sin \beta x d x$.
Solution. Let $I=\int e^{\alpha x} \sin \beta x d x$; as above, after applying the integration by parts formula twice, we will form an equation to find $I$. We have:

$$
\begin{gathered}
I=\int e^{\alpha x} \sin \beta x d x=\left\{\begin{array}{cc}
e^{\alpha x}=u, & d u=\alpha e^{\alpha x} d x \\
\sin \beta x d x=d v, & v=-\frac{1}{\beta} \cos \beta x
\end{array}\right\}= \\
=-\frac{e^{\alpha x}}{\beta} \cos \beta x+\frac{\alpha}{\beta} \int e^{\alpha x} \cos \beta x d x=\left\{\begin{array}{c}
e^{\alpha x}=u, d u=\alpha e^{\alpha x} \\
\cos \beta x d x=d v, \quad v=\frac{1}{\beta} \sin \beta x
\end{array}\right\}= \\
=-\frac{e^{\alpha x}}{\beta} \cos \beta x+\frac{\alpha}{\beta}\left[\frac{1}{\beta} e^{\alpha x} \sin \beta x-\frac{\alpha}{\beta} \int e^{\alpha x} \sin \beta x d x\right]= \\
=\frac{-\beta \cos \beta x+\alpha \sin \beta x}{\beta^{2}} \cdot e^{\alpha x}-\frac{\alpha^{2}}{\beta^{2}} I
\end{gathered}
$$

From the obtained equality, we find

$$
\int e^{\alpha x} \sin \beta x d x=\frac{\alpha \sin \beta x-\beta \cos \beta x}{\alpha^{2}+\beta^{2}} e^{\alpha x}+C
$$

Note the correct, sequential notations (in both cases $e^{\alpha x}=u$, although $e^{\alpha x} d x=d v$ could have been used) and that after the second integration, we obtained an integral similar to the one we are looking for.

Integration by parts can be used to integrate expressions containing high powers of the variable or certain irrationalities.