How to solve $y^{\prime \prime}(x)= a \cos (y)$? I was solving a physics questions and was stuck on this differential equation:  $y^{\prime \prime}(x)= a \cos (y)$, where $a$ is some constant.
I have no idea how to start. Please give a hint.
 A: Let $$z=\frac{dy}{dx},$$ then $$\frac{d^2y}{dx^2} = \frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx} = \frac{dz}{dy} z $$
Now we will apply this to your differential equation, 
$$ \frac{d^2y}{dx^2} = a \cos(y)$$
$$ z \frac{dz}{dy} = a \cos(y)$$
$$ \int z \ dz= \int a \cos(y)\ dy$$
$$ \frac12 z^2 =  a \sin(y) + C$$
Solve for $z$ and then integrate
$$ z = \sqrt{2}\sqrt{ a \sin(y) + C}$$
$$ \frac{dy}{dx} = \sqrt{2}\sqrt{ a \sin(y) + C}$$
$$ \int \frac{dy}{\sqrt{2}\sqrt{ a \sin(y) + C} } = \int dx $$
$$\boxed{ \int \frac{dy}{\sqrt{2}\sqrt{ a \sin(y) + C} } = x + D }$$
A: As a physicist, I have a sneaking suspicion that you've mis-copied your equation (which is very close to one of the most important physics equations).
But if not, here's your answer:
$$\left\{\left\{y(x)\to \frac{1}{2} \left(\pi -4 \text{am}\left(\frac{1}{2} \sqrt{\left(2
   a+c_1\right) \left(x+c_2\right){}^2}|\frac{4 a}{2
   a+c_1}\right)\right)\right\},\left\{y(x)\to \frac{1}{2} \left(4
   \text{am}\left(\frac{1}{2} \sqrt{\left(2 a+c_1\right) \left(x+c_2\right){}^2}|\frac{4
   a}{2 a+c_1}\right)+\pi \right)\right\}\right\}$$
