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 $f(t)=\sin (h bt)$
Now, $L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt \\= \int_{0}^{\infty} e^{-st} [\sin (h bt)] \ dt \\=\int_{0}^{\infty} \dfrac{e^{2st}e^{bt} \ dt}{2}-\int_{0}^{\infty} \dfrac{e^{-st}e^{-bt} \ dt}{2} \\=\dfrac{1}{2(s-b) }-\dfrac{1}{2(s+b)} \\=\dfrac{b}{(s^2-b^2) }$