SHM Solution W/O Small Angle!

for some time now, the small angle approximation has been a pet peeve of mine.
don't get me wrong, i greatly appreciate its uses! there are countless problems which would be unsolvable without it.
but i have always tried to get around using it, or when i have to, get approximate solutions.

i'm actually not positive on how much work has been done on solving the Harmonic Motion equation WITHOUT the small angle approximation. i do know that has no closed-form solution, but approximations have been done. and, in fact, i have one to present today!

it was actually not as much work as i thought it would be to get it into a form which could be approximated, but it definitely was more than is usually called for in a physics class i've been in.

so here goes. first, consider the free-body diagram of a swinging pendulum, and hopefully arrive at this equation:

-Lmg sinθ = Iα
where L is the length of the string, m is the mass of the object attached to it, g is the acceleration of gravity, I is the moment of inertia for the pendulum, and α is the rotational acceleration.
rearrange this and put it into the familiar differential equation form:

θ'' + c²sinθ = 0, where c² = Lmg/I

now let ω' = θ'' = -c²sinθ, then
dω/dt = -c²sinθ
also note that dθ/dt = ω. and we can combine the previous two equations to arrive at the following:

ω dω/dθ = -c²sinθ
ω dω = -c²sinθ dθ
∫ω dω = ∫-c²sinθ dθ + C₁
½ ω² = c² cosθ + C₁
ω² = 2c² cosθ + C₂

now to find C₂:
we know that at time t = 0, dθ/dt = ω(0) = 0, and θ(0) = θmax = θ₀
0² = 2c² cos(θ₀) + C₂
C₂ = -2c² cos(θ₀)

there so now we have:
ω² = 2c² cosθ - 2c² cos(θ₀)
... skipping a few steps now in the interest of space:
dθ/dt = ± c(√2)√[ cos(θ) - cos(θ₀) ]
this D.E. is separable and leads to the following integral:

∫dθ/√[ cos(θ) - cos(θ₀) ] = ± c(√2) t + C₃
unfortunately, this is an elliptic integral and cannot be solved explicitly. so i will just call it I and solve for t:
I = ± c(√2) t
t = I/(± c(√2)) + C₄.

you can, however, approximate this integral with a Taylor Series. i'll give the first few terms here. but note that  the rest can be found by plugging the integral into Wolfram Alpha's solver. and in the interest of clarity, i will post a picture instead of writing the terms out:
∫dx/√[ cos(x) - a ] ≈ ...

http://www.wolframalpha.com/input/?i=integrate+dx/(cosx+-+a)^(1/2)

it is nice to see this solution compared to the solution of the equation WITH the small angle approximation. we can write an expression for t of this equation as such:

t = arccos(θ/θ₀)/c

(much simpler than the other form!)

i would have liked to graph these two functions and see how accurately the small angle approximation approximates the actual function, but i couldn't find a grapher that was accurate enough, nor enough terms of the integral to get it sufficiently accurate.
i will, however, tell you that it actually approximates the true solution VERY well, and is alright in my book.
if you are interested in plotting or graphing these values, the Wolfram Alpha site will let you get copy-able text to paste into your software by clicking on the picture.
i found this website convenient for quick graphing for anyone who wants it:
(not very accurate though)

http://www.livephysics.com/ptools/online-function-grapher.php