Solving $ \frac{d^2}{dt^2}\theta = k\sin\theta$ for $t$ Let $k$ be any constant, given $ \theta(0)=0$, $ \frac d{dt}\theta=0$ when $t=0$, t = ? if $ \theta = \frac \pi 2$ where $t$ represents time.
$ \frac{d^2}{dt^2}\theta = k\sin\theta $
How would I solve this problem in the simplest manner? This can be modeled with large angle pendulum or falling stick (of unifom thickness) falling from unstable equilibrium.

Since we can calculate the time taken by the blob from $ \pi/2  \text{ to } 0 $ ( or $ \pi $ ) shouldn't we able to calculate the time theoretically? Correct me if I'm mistaken.
Let the parameter be  $ \theta(0)=\pi/2$, $ \frac {d\theta}{dt}=0$ when $t=0$, t = ? if $ \theta = \pi \text{( or 0) } $
 A: The solution of the initial value problem $\theta'' = k \sin \theta$, $\theta(0)= \pi/2$, $\theta'(0)=0$, is given implicitly (for $\pi/2 \le \theta(t) \le 3\pi/2$, i.e. the first swing of the pendulum) by
$$ \int _{\pi/2 }^{\theta \left( t \right) }\!{\frac {1}{\sqrt {-2\,k
\cos \left( s \right) }}}\ {ds}=t$$
The time to go from $\theta = \pi/2$ to $\theta=\pi$ is thus
$$ \int_{\pi/2}^\pi \frac{1}{\sqrt{-2k\cos(s)}}\ ds = \frac{1}{\sqrt{k}} \int_{\pi/2}^\pi \frac{1}{\sqrt{-2\cos(s)}}\ ds$$
That last integral is non-elementary: its approximate value is $1.854074677$, and it can be expressed as ${\rm EllipticK}(1/\sqrt{2})$ in the convention used by Maple.  Wolfram Alpha calls it $K(1/2)$.  It can also be written as $\dfrac{\pi^{3/2}}{2 \Gamma(3/4)^2}$.
A: As a complement to the above multiplying both sides by $\frac{d \theta}{dt}dt=d\theta$ we obtain
$$\frac{d^2 \theta}{dt^2}\frac{d \theta}{dt}dt=k\sin\theta d\theta$$
$$\frac{1}{2}\left(\frac{d\theta}{dt}\right)^2=k\int_{\frac{\pi}{2}}^{\theta}\sin\theta d\theta$$
also referred to as the energy integral.
$$\frac{1}{2}\left(\frac{d\theta}{dt}\right)^2=-k\cos\theta$$
Now solve for $dt$ and integrate from $\frac{\pi}{2}$ to $\theta$
This is a general trick of lowering the order of equations of the form
$$z''=f(z)$$
