I am using this page to do two things: (i) answer questions posed on the Quora website, and (ii) test MathJax in doing so.
As pointed out by Dzung Nuyen, this is a form of the equation of motion for a simple pendulum (of length $L$, accelerated by gravity $g$, where $K = g/L$). For small values of $\theta$ for which $\sin{\theta} \approx \theta$, its solutions are the well known simple harmonic solutions given by Dzung Nuyen. However, you asked how to solve the first-order ODE represented by the first integral of the equation of motion, presumably for all values of $\theta$ satisfying $\cos{\theta} > \cos{\theta_0}$. To do this, begin by separating variables and writing it as
$$\dfrac{d\theta}{\sqrt{2(\cos{\theta} - \cos{\theta_0})}} = -\sqrt{K}\,dt.$$
Make the trig substitutions $\cos{\theta} = 1 - 2\sin^2{(\theta/2)}$ and $\cos{\theta_0} = 1 - 2\sin^2{(\theta_0/2)}$ to bring this to the form
$$\dfrac{d\theta}{\sqrt{4[\sin^2{(\theta_0/2)} - \sin^2{(\theta/2)}]}} = -\sqrt{K}\,dt,$$
or, equivalently
$$\dfrac{d(\theta/2)}{\sqrt{1 - \csc^2{(\theta_0/2)}\sin^2{(\theta/2)}}} = -\sqrt{K}\,\sin^2{(\theta_0/2)}\,dt,$$
Siddhant Das has presented a succinct solution to the problem, impeccable in every detail, as we have come to expect of him. Just excellent work. I thought I had found an alternative form of the solution by approaching the reduction in a slightly different way. As it turns out, what I end up with is something Siddhant found early on in his procedure, but chose not to emphasize. Perhaps it is still worth presenting as an alternative approach; as usual, I apologize for being overly pedantic. I do this in part to have a good record of what was done, so I never have to wonder about how I got from one step to the next.
\[\begin{align}\mathscr{L}[\sin{(\ln{t})}] &= \int_0^\infty\,\sin{(\ln{t})}\,e^{-st}\,dt\\
&= \frac{1}{s}\int_0^\infty\,\sin{\left[\ln{\left(\frac{u}{s}\right)}\right]}\,e^{-u}\,du, \qquad {\text{(substituted $u=st$)}}\\
&= \frac{1}{s}\int_0^\infty\,\sin{ (\ln{u}-\ln{s} )}\,e^{-u}\,du\\
&= \frac{1}{s}\int_0^\infty\,[\sin{(\ln{u})}\cos{(\ln{s})}-\cos{(\ln{u})}\sin{(\ln{s})}]\,e^{-u}\,du, \qquad {\text{(used trig identity)}}\\
&= \frac{\cos{(\ln{s})}}{s}\int_0^\infty\,\sin{(\ln{u})}\,e^{-u}\,du-\frac{\sin{(\ln{s})}}{s}\int_0^\infty\,\cos{(\ln{u})}\,e^{-u}\,du\end{align}\]
Under the integrals, replace the sine and cosine by their complex exponential equivalents: $\sin{x}=(e^{ix}-e^{-ix})/2i$, and $\cos{x} = (e^{ix} + e^{-ix})/2$, so that $\sin{(\ln{u})}=(e^{i\ln{u}} - e^{-i\ln{u}})/2i = (u^i - u^{-i})/2i$ and $\cos{(\ln{u})}=(e^{i\ln{u}} + e^{-i\ln{u}})/2 = (u^i + u^{-i})/2$. This brings the Laplace transform to
\[\begin{align}\mathscr{L}[\sin{(\ln{t})}] &= \frac{\cos{(\ln{s})}}{2is}\int_0^\infty\,(u^i - u^{-i})\,e^{-u}\,du-\frac{\sin{(\ln{s})}}{2s}\int_0^\infty\,(u^i + u^{-i})\,e^{-u}\,du\end{align}\]
Each of the remaining integrals can now be recognized as either $\Gamma(1+i) = \int_0^\infty\,u^i\,e^{-u}\,du$, or its complex conjugate $\Gamma(1-i)= [\Gamma(1+i)]^* = \int_0^\infty\,u^{-i}\,e^{-u}\,du$, see WolframAlpha. It follows that $\Gamma(1+i)-\Gamma(1-i) = 2i{\text{Im}}{[\Gamma(1+i)]}$ and $\Gamma(1+i) + \Gamma(1 - i) = 2{\text{Re}}{[\Gamma(1+i)]}$, so the transform reduces to
\[\begin{align}\mathscr{L}[\sin{(\ln{t})}] &= \frac{\cos{(\ln{s})}}{2is}\,\left(2i{\text{Im}}{[\Gamma(1+i)]}\right) - \frac{\sin{(\ln{s})}}{2s}\,\left(2{\text{Re}}{[\Gamma(1+i)]}\right)\\
&= \frac{{\text{Im}}{[\Gamma(1+i)]}\,\cos{(\ln{s})} - {\text{Re}}{[\Gamma(1+i)]}\,\sin{(\ln{s})}}{s}\end{align}\]
From WolframAlpha, the gamma function evaluated at $1+i$ is, to six decimal places, $\Gamma(1+i) \approx 0.498016 - 0.154950\,i$, allowing the transform to be written as
\[\begin{align}\mathscr{L}[\sin{(\ln{t})}] =-\left[\,\frac{0.154950\,\cos{(\ln{s})} + 0.498016\,\sin{(\ln{s})}}{s}\,\right]\end{align}\]
As pointed out by Dzung Nuyen, this is a form of the equation of motion for a simple pendulum (of length $L$, accelerated by gravity $g$, where $K = g/L$). For small values of $\theta$ for which $\sin{\theta} \approx \theta$, its solutions are the well known simple harmonic solutions given by Dzung Nuyen. However, you asked how to solve the first-order ODE represented by the first integral of the equation of motion, presumably for all values of $\theta$ satisfying $\cos{\theta} > \cos{\theta_0}$. To do this, begin by separating variables and writing it as
$$\dfrac{d\theta}{\sqrt{2(\cos{\theta} - \cos{\theta_0})}} = -\sqrt{K}\,dt.$$
Make the trig substitutions $\cos{\theta} = 1 - 2\sin^2{(\theta/2)}$ and $\cos{\theta_0} = 1 - 2\sin^2{(\theta_0/2)}$ to bring this to the form
$$\dfrac{d\theta}{\sqrt{4[\sin^2{(\theta_0/2)} - \sin^2{(\theta/2)}]}} = -\sqrt{K}\,dt,$$
or, equivalently
$$\dfrac{d(\theta/2)}{\sqrt{1 - \csc^2{(\theta_0/2)}\sin^2{(\theta/2)}}} = -\sqrt{K}\,\sin^2{(\theta_0/2)}\,dt,$$
Siddhant Das has presented a succinct solution to the problem, impeccable in every detail, as we have come to expect of him. Just excellent work. I thought I had found an alternative form of the solution by approaching the reduction in a slightly different way. As it turns out, what I end up with is something Siddhant found early on in his procedure, but chose not to emphasize. Perhaps it is still worth presenting as an alternative approach; as usual, I apologize for being overly pedantic. I do this in part to have a good record of what was done, so I never have to wonder about how I got from one step to the next.
\[\begin{align}\mathscr{L}[\sin{(\ln{t})}] &= \int_0^\infty\,\sin{(\ln{t})}\,e^{-st}\,dt\\
&= \frac{1}{s}\int_0^\infty\,\sin{\left[\ln{\left(\frac{u}{s}\right)}\right]}\,e^{-u}\,du, \qquad {\text{(substituted $u=st$)}}\\
&= \frac{1}{s}\int_0^\infty\,\sin{ (\ln{u}-\ln{s} )}\,e^{-u}\,du\\
&= \frac{1}{s}\int_0^\infty\,[\sin{(\ln{u})}\cos{(\ln{s})}-\cos{(\ln{u})}\sin{(\ln{s})}]\,e^{-u}\,du, \qquad {\text{(used trig identity)}}\\
&= \frac{\cos{(\ln{s})}}{s}\int_0^\infty\,\sin{(\ln{u})}\,e^{-u}\,du-\frac{\sin{(\ln{s})}}{s}\int_0^\infty\,\cos{(\ln{u})}\,e^{-u}\,du\end{align}\]
Under the integrals, replace the sine and cosine by their complex exponential equivalents: $\sin{x}=(e^{ix}-e^{-ix})/2i$, and $\cos{x} = (e^{ix} + e^{-ix})/2$, so that $\sin{(\ln{u})}=(e^{i\ln{u}} - e^{-i\ln{u}})/2i = (u^i - u^{-i})/2i$ and $\cos{(\ln{u})}=(e^{i\ln{u}} + e^{-i\ln{u}})/2 = (u^i + u^{-i})/2$. This brings the Laplace transform to
\[\begin{align}\mathscr{L}[\sin{(\ln{t})}] &= \frac{\cos{(\ln{s})}}{2is}\int_0^\infty\,(u^i - u^{-i})\,e^{-u}\,du-\frac{\sin{(\ln{s})}}{2s}\int_0^\infty\,(u^i + u^{-i})\,e^{-u}\,du\end{align}\]
Each of the remaining integrals can now be recognized as either $\Gamma(1+i) = \int_0^\infty\,u^i\,e^{-u}\,du$, or its complex conjugate $\Gamma(1-i)= [\Gamma(1+i)]^* = \int_0^\infty\,u^{-i}\,e^{-u}\,du$, see WolframAlpha. It follows that $\Gamma(1+i)-\Gamma(1-i) = 2i{\text{Im}}{[\Gamma(1+i)]}$ and $\Gamma(1+i) + \Gamma(1 - i) = 2{\text{Re}}{[\Gamma(1+i)]}$, so the transform reduces to
\[\begin{align}\mathscr{L}[\sin{(\ln{t})}] &= \frac{\cos{(\ln{s})}}{2is}\,\left(2i{\text{Im}}{[\Gamma(1+i)]}\right) - \frac{\sin{(\ln{s})}}{2s}\,\left(2{\text{Re}}{[\Gamma(1+i)]}\right)\\
&= \frac{{\text{Im}}{[\Gamma(1+i)]}\,\cos{(\ln{s})} - {\text{Re}}{[\Gamma(1+i)]}\,\sin{(\ln{s})}}{s}\end{align}\]
From WolframAlpha, the gamma function evaluated at $1+i$ is, to six decimal places, $\Gamma(1+i) \approx 0.498016 - 0.154950\,i$, allowing the transform to be written as
\[\begin{align}\mathscr{L}[\sin{(\ln{t})}] =-\left[\,\frac{0.154950\,\cos{(\ln{s})} + 0.498016\,\sin{(\ln{s})}}{s}\,\right]\end{align}\]