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$\int e^{-2x}sin5xdx=-\frac{1}{5}e^{-2x}cos5x-\frac{2}{5}\int e^{-2x}cos5xdx=\\\\=-\frac{1}{5}e^{-2x}cos5x-\frac{2}{25}e^{-2x}sin5x-\frac{4}{25}\int e^{-2x}sin5xdx\\\\\\ \frac{29}{25}\int e^{-2x}sin5xdx=-\frac{1}{5}e^{-2x}cos5x-\frac{2}{25}e^{-2x}sin5x\\\\\\\int e^{-2x}sin5xdx=-\frac{5}{29}e^{-2x}cos5x-\frac{2}{29}e^{-2x}sin5x+C$
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$\int e^{-2x}sin5xdx=-\frac{1}{5}e^{-2x}cos5x-\frac{2}{5}\int e^{-2x}cos5xdx\\\\u=e^{-2x}=\ \textgreater \ du=-2e^{-2x}dx\\dv=sin5xdx=\ \textgreater \ v=-\frac{1}{5}cos5x\\\\\int e^{-2x}cos5xdx=\frac{1}{5}e^{-2x}sin5x+\frac{2}{5}\int e^{-2x}sin5xdx\\\\p=e^{-2x}=\ \textgreater \ dp=-2e^{-2x}\\dq=cos5xdx=\ \textgreater \ q=\frac{1}{5}sin5x$