$$T = 4\sqrt{\frac{l}{2g}} \int_0^x{\frac{dy}{\sqrt{\cos(y)-\cos(x)}}}$$
So do i just have to find a way so that the denominator does not result in 0? I am to later evaluate it with x = 51. Which leads me to trouble because I get a 0 in the denominator at x = 51.
EDIT:the problem is that I need to figure out how to evaluate this integral and get an answer without actually changing the integrand itself like i did the first time
Second Edit: had to fix question
Third edit: Potential Solution
x is a constant therefore, cos(x) is treated as a constant.
$cos(y) - cos(x) = t^2$
$-sin(y)dy = 2t dt$
$dy = -\frac{2t}{sin(y)}dt$
$\int_0^x{-\frac{2t}{sin(y)} * \frac{1}{t}}dt$ = $\int_0^x{-\frac{2}{sin(y)}dt}$ = $\frac{-2t}{sin(y)}$ = $\frac{-2\sqrt{(cos(y)-cos(x))}}{sin(y)}$ from y = 0 to x. X is later given as 51 so the solution would be just plug and solve from here?
This can be written in terms of an elliptic integral of the first type using the identity $$ 2\sin^2{\alpha} =1-\cos{(2\alpha)}$$
$$T=4\sqrt{\frac{l}{2g}}\int_{0}^{x}\frac{dy}{\sqrt{\cos{y}-\cos{x}}}=2\sqrt{\frac{l}{g}}\int_{0}^{x}\frac{dy}{\sqrt{\sin^2{(x/2)}-\sin^2{(y/2)}}}$$ Now the substitution $\sin{\xi}=\sin{(y/2)}/\sin{(x/2)}$ yields to $$T = 4\sqrt{\frac{l}{g}}\int_{0}^{\pi/2}\frac{d\xi}{\sqrt{1-(\sin{(x/2))^2\sin^2{\xi}}}}$$ This integral does not have any primitive and must be solved numerically.
$\textbf{Adendum}$
In order to compute $dy(\xi)$ apply the chain rule to $dy(\xi)=\frac{dy}{d\xi}d\xi$. To obtain this derivative, derive the equation with the proposed change of variables with respect to $\xi$ as it follows: $$\sin{(x/2)}\cos{\xi}=\frac{1}{2}\cos{(y/2)}\frac{dy}{d\xi}=\frac{1}{2}\sqrt{1-\sin^2{(y/2)}}\frac{dy}{d\xi}=\frac{1}{2}\sqrt{1-(\sin{(x/2)}^2\sin^2{\xi}}\frac{dy}{d\xi}$$ and solve for $\frac{dy}{d\xi}$