Question:

Let $I = \int e^{ax} \sin bx dx$. Integrate by parts taking $e^{ax}$ as the first function, we get $I = \frac{-e^{ax} \cos bx}{b} - \int a e^{ax} \left( \frac{-\cos bx}{b} \right) dx$. On integrating the second term by parts again, we get $I = -\frac{1}{b} e^{ax} \cos bx + \frac{a}{b} \left[ \frac{1}{b} e^{ax} \sin bx - \int \frac{a}{b} e^{ax} \sin bx dx \right] \Rightarrow I = -\frac{1}{b} e^{ax} \cos bx + \frac{a}{b^2} e^{ax} \sin bx - \frac{a^2}{b^2} I \Rightarrow \left( 1 + \frac{a^2}{b^2} \right) I = \frac{e^{ax}}{b^2} (a \sin bx - b \cos bx) + C \Rightarrow I = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) + C$. Similarly we can show that $\int e^{ax} \cos bx dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) + C$.