Consider the pendulum problem
$\frac{d^2x}{dt^2}+\sin(x)=0$
$\frac{dx}{dt}(0)=v_0=0$
$x(0)=x_0$
Show that the period $T=\displaystyle2\int_{-x_o}^{x_0}\frac{dx}{\sqrt{\cos(x)-\cos(x_0)}}$
Why do I get: $T=\displaystyle2\int_{-x_o}^{x_0}\frac{dx}{\sqrt{2(\cos(x)-\cos(x_0)})}$, is the question wrong, can you check my steps please !
$\frac{d^2x}{dt^2}+\sin(x)=0\implies\frac{dx}{dt}(\frac{d^2x}{dt^2}+\sin(x))=0$
but $\frac{dx}{dt}(\frac{d^2x}{dt^2}+\sin(x))=\frac{d}{dt}\Big(\frac12(\frac{dx}{dt})^2-\cos(x)\Big)$
Hence
$\frac12(\frac{dx}{dt})^2-\cos(x)=C$,$\qquad$$\text{C is Constant}$
With the initial value I obtain;
$C=-\cos(x_0)$ Therefore;
$\frac12(\frac{dx}{dt})^2=\cos(x)-\cos(x_0)\implies (\frac{dx}{dt})=-\sqrt{2(\cos(x)-\cos(x_0))}$
Sign is negative because $x$ is decreasing, with separation of variables it becomes to:
$\displaystyle\int_{0}^{T/4}dt=-\int_{x_0}^{0}\frac{dx}{\sqrt{2(\cos(x)-\cos(x_0))}}\implies T=2\int_{-x_o}^{x_0}\frac{dx}{\sqrt{2(\cos(x)-\cos(x_0)})}$