$ Consider\\[2ex] \displaystyle \int_0^{\infty{}}e^{-st}\ \sin \ \left(\dfrac{t}{2}\right)\ \sin h{\ \left(\dfrac{\sqrt{3}t}{2}\right)}\ dt\ =\ L\left[\begin{array}{l}\ \sin {\left(\dfrac{t}{2}\right)\ \sin h{\ \left(\dfrac{\sqrt{3}t}{2}\right)}}\end{array}\right] \\[7ex] Now,\\[2ex] \sin {\left(\dfrac{t}{2}\right)\ \sin h{\ \left(\dfrac{\sqrt{3}t}{2}\right)}}=\sin \ \left(\dfrac{t}{2}\right)\ \left(\dfrac{e^{\dfrac{\sqrt{3}t}{2}}-e^{\dfrac{-\sqrt{3}t}{2}}\ }{2}\right) \\[9ex] By \ First \ shifting \ theorem,\\[2ex] \begin{align*} \displaystyle L\ \left[e^{\dfrac{\sqrt{3}t}{2}}\ \sin {\left(\dfrac{t}{2}\right)}\right]&=\ \dfrac{\dfrac{1}{2}}{{\left(\ s\ –\ \dfrac{\sqrt{3}}{2}\right)}^2}+\dfrac{1}{4} \\[2ex] \displaystyle &=\ \dfrac{\dfrac{1}{2}}{s^2-\sqrt{3}s+1} \\[2ex] \displaystyle &=\ \dfrac{\dfrac{1}{2}}{{\left(\ s+\dfrac{\sqrt{3}}{2}\right)}^2+\dfrac{1}{4}}\ \\[2ex] \displaystyle &=\ \dfrac{\dfrac{1}{2}}{s^2+\sqrt{3}s+\dfrac{3}{4}+\dfrac{1}{4}} \\[2ex] \displaystyle &=\ \dfrac{\dfrac{1}{2}}{s^2+\sqrt{3}s+1} \\[2ex] \end{align*} $

$\begin{align*} \displaystyle \therefore{}\ \ L\left[\begin{array}{l}\sin {\left(\dfrac{t}{2}\right)\sin h{\ \left(\dfrac{\sqrt{3}t}{2}\right)}}\end{array}\right]\ &=\ \dfrac{1}{2}\ \left[\dfrac{\dfrac{1}{2}}{{(s}^2+1)-\sqrt{3}s}-\ \dfrac{\dfrac{1}{2}}{{(s}^2+1)+\sqrt{3}s}\right] \\[2ex] \displaystyle &=\ \dfrac{1}{2}\ .\ \dfrac{1}{2}\ \left[\dfrac{s^2+1+\sqrt{3}s\ –\ \left(s^2+1-\sqrt{3}s\right)}{{{(s}^2+1)}^2-\ 3s^2}\right] \\[2ex] \displaystyle &=\ \dfrac{1}{4}\ \left(\dfrac{2\sqrt{3}s}{s^4-\ s^2+1\ }\right) \\[2ex] \displaystyle &=\ \dfrac{\sqrt{3}}{2}\ \left(\dfrac{s}{s^4-\ s^2+1\ }\right) \\[2ex] \end{align*}\\ \displaystyle \therefore{}\ \int_0^{\infty{}}e^{-st}\ \sin \ \left(\dfrac{t}{2}\right)\sin h{\ \left(\dfrac{\sqrt{3}t}{2}\right)}\ dt\ =\ \dfrac{\sqrt{3}}{2}\ \left(\dfrac{s}{s^4-\ s^2+1\ }\right) \\[5ex] Putting s=1,\\[4ex] $

$\begin{align*} \displaystyle \int_0^{\infty{}}e^{-t}\ \sin \ \left(\dfrac{t}{2}\right)\sin h{\ \left(\dfrac{\sqrt{3}t}{2}\right)}\ dt\ &=\ \dfrac{\sqrt{3}}{2}\ \left(\dfrac{1}{1^4-\ 1^2+1\ }\right) \\[2ex] &=\ \dfrac{\sqrt{3}}{2}\\[2ex] \end{align*}$

$\displaystyle \therefore{}\ \int_0^{\infty{}}e^{-t}\ \sin \ \left(\dfrac{t}{2}\right)\sin h{\ \left(\dfrac{\sqrt{3}t}{2}\right)}\ dt\ =\ \dfrac{\sqrt{3}}{2} \\[2ex] $