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The value of the integral \int_{0}^{ \infty }{e^{-2t} sin t} dt= is

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The value of the integral $\int_{0}^{ \infty }{e^{-2t} sin t} dt=$ \int_{0}^{ \infty }{e^{-2t} sin t} dt= is

$\frac{1}{5}$ \frac{1}{5}

$\frac{1}{3}$ \frac{1}{3}

$\frac{1}{2}$ \frac{1}{2}

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