Answer
$\dfrac{b}{s^2+b^2} $
Work Step by Step
The Laplace Transform can be written as:
$L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt $
We are given that $f(t)=\sin (bt)$
Now, $L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt \\= \int_{0}^{\infty} e^{-st} [\sin (bt)] \ dt \\=\lim\limits_{n \to \infty} \int_{0}^{\infty} e^{-st} [\sin (bt)] \ dt\\=\lim\limits_{n \to \infty} [ e^{-st} [-\sin (bt)-b \cos (bt)]_{0}^{\infty} \ dt\\=\dfrac{b}{s^2+b^2} $
