The equation of motion of pendulum 1-1 can be derived as shown in the equation (1) below. In this chapter, we would like to obtain the exact solution of the pendulum analyzing in the form of $sin \theta$ without applying the approximation, while in the Chapter 1, analyzing in the approximated form as ''$\sin\theta\approx\theta$'' when ''$\theta$'' is small.
\begin{eqnarray}
ml\frac{d^2\theta}{dt^2}=-mgsin\theta
\end{eqnarray}
Dividing both sides of the equation by ''$ml$'', multiplying ''$d\theta/dt$'' and transposing the right side to the left side, the equation (1) can be converted to the eqation (2).
\begin{eqnarray}
\frac{d^2\theta}{dt^2}\frac{d\theta}{dt}+\frac{g}{l}\frac{d\theta}{dt}sin\theta=0
\end{eqnarray}
The equation (2) can be converted to the equation (3). Accordingly, the equation (4) can be obtained.
\begin{eqnarray}
\frac{d}{dt}\{\frac{1}{2}(\frac{d\theta}{dt})^2-\omega_n^2cos\theta\}=0
\end{eqnarray}
\begin{eqnarray}
\frac{1}{2}(\frac{d\theta}{dt})^2-\omega_n^2cos\theta=E
\end{eqnarray}
This physically represents the law of the conservation of energy. The entire energy can be represented as {$ml^2E+mgl$}.
Here, assuming that ''$d\theta/dt = 0$'' when in ''$\theta=A_0$'', the equation (5) can be derived from the equation (4).
\begin{eqnarray}
E=-\omega_n^2cosA_0
\end{eqnarray}
Accordingly the equation (4) can also be converted to the equation (6).
\begin{eqnarray}
\frac{1}{2}(\frac{d\theta}{dt})^2-\omega_n^2(cos\theta-cosA_0)=0
\end{eqnarray}
In other words, the equation (6) can be converted to the equation (7).
\begin{eqnarray}
\frac{d\theta}{dt}=\pm\sqrt{\frac{2g(cos\theta-cosA_0)}{l}}
\end{eqnarray}
Separating this to the integration with regard to ''$\theta$'' & ''$t$'', as the time for ''$\theta$'' to swing from ''0'' to ''$A_0$'' is ''$T/4$'', the equation (8) can be obtained.
\begin{eqnarray}
\int_0^{A_0} \frac{1}{\sqrt{cos\theta-cosA_0}} d\theta=\int_0^{\frac{T}{4}} \sqrt{\frac{2g}{l}} dt=\sqrt{\frac{2g}{l}}\frac{T}{4}
\end{eqnarray}
Here, based on the assumption of the equation (9), inside the denominator of the left side of the equation (8) can be obtained as shown in the equation (10).
\begin{eqnarray}
sin\frac{\theta}{2}=sin\frac{A_0}{2}sin\phi
\end{eqnarray}
\begin{eqnarray}
cos\theta-cosA_0=(1-2sin^2\frac{\theta}{2})-(1-2sin^2\frac{A_0}{2})
&=&2sin^2\frac{A_0}{2}cos^2\phi
\end{eqnarray}
In addition, differentiating the both sides of the equation (9), it can be converted to the equation (11)
\begin{eqnarray}
\frac{1}{2}cos\frac{\theta}{2} d\theta=sin\frac{A_0}{2}cos\phi\;d\phi
\end{eqnarray}
The equation (12) can be obtained from the equation (11).
\begin{eqnarray}
d\theta&=&\frac{2sin\frac{A_0}{2}cos\phi}{cos\frac{\theta}{2}}d\phi \nonumber \\
&=&\frac{2sin\frac{A_0}{2}cos\phi}{\sqrt{1-sin^2\frac{A_0}{2}sin^2\phi}}\;d\phi
\end{eqnarray}
Paying attention that from the equation (9), it becomes ''$\phi= 0$'' & ''$\phi= \pi/2$''
in case of ''$\theta= 0$'' & ''$\theta= A_0$'' respectively, and substituting the equations (10) & (12) to
the equation (8), the equation (13) can be obtained.
\begin{eqnarray}
T&=&4\sqrt\frac{l}{2g}\int_0^{A_0}\frac{1}{\sqrt{cos\theta-cosA_0}}d\theta \nonumber \\
&=&4\sqrt\frac{l}{2g}\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{2sin^2\frac{A_0}{2}cos^2\phi}}\frac{2sin\frac{A_0}{2}cos\phi}{\sqrt{1-sin^2\frac{A_0}{2}sin^2\phi}}d\phi \nonumber \\
&=&4\sqrt{\frac{l}{g}}\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1-sin^2\frac{A_0}{2}sin^2\phi}}d\phi
\end{eqnarray}
Here, the following equation (14) is called as elliptic integral of the first kind\index{elliptic integral of the first kind}, and Taylor series expansion can be made in terms of inside of the integral as shown in the equation (15).
\begin{eqnarray}
K(m)=\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1-m sin^2\theta}}d\theta
\end{eqnarray}
\begin{eqnarray}
\frac{1}{\sqrt{1-msin^2\theta}}&=&1+\frac{1}{2}msin^2\theta+\frac{3}{8}m^2sin^4\theta+\frac{15}{48}m^3sin^6\theta+... \nonumber \\
&=&\sum_{i=0}^\infty \frac{(2n-1)!!}{(2n)!!}m^nsin^{2n}\theta
\end{eqnarray}
As this sequence of number makes the uniform convergence in the range of ''$m < 1$'', the equation (16) can be obtained by applying the termwise integration. For your note, the mark '' !!'' means that the alternate factorial is to be applied.
\begin{eqnarray}
K(m)&=&\int_0^{\frac{\pi}{2}}\sum_{i=0}^\infty \frac{(2n-1)!!}{(2n)!!}m^nsin^{2n}\theta d\theta \nonumber \\
&=&\sum_{i=0}^\infty \frac{(2n-1)!!}{(2n)!!}m^n \int_0^{\frac{\pi}{2}}sin^{2n}\theta d\theta \nonumber \\
&=&\frac{\pi}{2}\{\sum_{i=0}^\infty [\frac{(2n-1)!!}{(2n)!!}]^2m^n\}
\end{eqnarray}
With the use of ''$K(m)$'', the equation (17) can be obtained from the equation (13).
\begin{eqnarray}
T&=&4\sqrt{\frac{l}{g}}\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1-sin^2\frac{A_0}{2}sin^2\phi}}d\phi \nonumber \\
&=&4\sqrt{\frac{l}{g}}K(sin^2\frac{A_0}{2}) \nonumber \\
&=&2\pi\sqrt{\frac{l}{g}}\{\sum_{i=0}^\infty [\frac{(2n-1)!!}{(2n)!!}]^2sin^{2n}\frac{A_0}{2}\} \nonumber \\
&=&2\pi\sqrt{\frac{l}{g}}\{1+\frac{1}{4}sin^2\frac{A_0}{2}+\frac{9}{64}sin^4\frac{A_0}{2}+\frac{225}{2304}sin^6\frac{A_0}{2}+\frac{10125}{147456}sin^8\frac{A_0}{2}...\}
\end{eqnarray}
As shown the above, the period of the pendulum, ''$T$'' is a function of the displacement, ''$A_0$'' exactly, and based on the assumption with ''$A_0=0$'', it corresponds with the equation (12) in 1-1.
This is the exact solution of the pendulum.