Integral of e^(-2x)*sin(5x) dx

1. Use integration by parts, noting that the integrand eventually repeats itself.

   1. For the integrand $e^{- 2 x} \sin{\left(5 x \right)}$:

      Let $u{\left(x \right)} = \sin{\left(5 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = e^{- 2 x}$.

      Then $\int e^{- 2 x} \sin{\left(5 x \right)}\, dx = - \int \left(- \frac{5 e^{- 2 x} \cos{\left(5 x \right)}}{2}\right)\, dx - \frac{e^{- 2 x} \sin{\left(5 x \right)}}{2}$.

   2. For the integrand $- \frac{5 e^{- 2 x} \cos{\left(5 x \right)}}{2}$:

      Let $u{\left(x \right)} = - \frac{5 \cos{\left(5 x \right)}}{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{- 2 x}$.

      Then $\int e^{- 2 x} \sin{\left(5 x \right)}\, dx = \int \left(- \frac{25 e^{- 2 x} \sin{\left(5 x \right)}}{4}\right)\, dx - \frac{e^{- 2 x} \sin{\left(5 x \right)}}{2} - \frac{5 e^{- 2 x} \cos{\left(5 x \right)}}{4}$.

   3. Notice that the integrand has repeated itself, so move it to one side:

      $$\frac{29 \int e^{- 2 x} \sin{\left(5 x \right)}\, dx}{4} = - \frac{e^{- 2 x} \sin{\left(5 x \right)}}{2} - \frac{5 e^{- 2 x} \cos{\left(5 x \right)}}{4}$$

      Therefore,

      $$\int e^{- 2 x} \sin{\left(5 x \right)}\, dx = - \frac{2 e^{- 2 x} \sin{\left(5 x \right)}}{29} - \frac{5 e^{- 2 x} \cos{\left(5 x \right)}}{29}$$

2. Now simplify:

   $$- \frac{\left(2 \sin{\left(5 x \right)} + 5 \cos{\left(5 x \right)}\right) e^{- 2 x}}{29}$$

3. Add the constant of integration:

   $$- \frac{\left(2 \sin{\left(5 x \right)} + 5 \cos{\left(5 x \right)}\right) e^{- 2 x}}{29}+ \mathrm{constant}$$

The answer is:

$$- \frac{\left(2 \sin{\left(5 x \right)} + 5 \cos{\left(5 x \right)}\right) e^{- 2 x}}{29}+ \mathrm{constant}$$