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I suck at integrals. Here on this Wiki list, it says that $$\int \mathrm{e}^{cx} \sin bx \,\mathrm{d}x = \frac{\mathrm{e}^{cx}}{b^2 + c^2}(c \sin bx - b \cos bx),$$ which is easy to check – but how could I have derived that?

k.stm
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    Differentiating the RHS is a fully legitimate way to justify the identity, though it assumes you know the anti-derivative ahead of time. To derive it from scratch, the standard method is repeated integration by parts. – David H Jun 14 '15 at 15:34
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    $\int e^{cx}\sin(bx)=\Im\int e^{cx+ ibx}$ is fast – tired Jun 14 '15 at 15:34

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Just write $\sin(bx)$ as $\text{Im}\left(e^{ibx}\right)$, so that the LHS is just the imaginary part of $\frac{1}{c+ib} e^{(c+ib)x}$.

Jack D'Aurizio
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If you wanted to acquire the result without actually differentiating the given integral, then you can use integration by parts twice. For the first application, set $u=\sin (bx)$ and $dv=e^{cx}$. For the second application, set $u= \cos (bx)$ and $dv=e^{cx}$.

Trogdor
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