Find the domain of $F(x)$, defined by$$F(x)=-\int_0^{x}{\log(\cos t)}dt$$
My attempt:
$$f(x)=\log(\cos x), \ D(f(x))=\left(-\frac\pi{2},\frac\pi{2}\right)$$
So I did this:
$$-\int_{0}^{-\frac\pi{2}}{\log(\cos t)}dt =\int_{-\frac\pi{2}}^{0}{\log(\cos t)}dt$$
I am having problems with determining whether this integral converges or not.
I tried using Taylor: $\int_{-\frac\pi{2}}^{0}{\log(t+\frac{\pi}2)}dt$, but I don't know how to go on.
I then tried substituting $x=t+\frac{\pi}2$ so: $\int_{0}^{\frac\pi{2}}{\log(\cos x)}dx$ but I don't know how to go on either.
EDIT: New attempt:
I continued from here: $$\int_{-\frac\pi{2}}^{0}{\log\left(t+\frac{\pi}2\right)}dt = \int_{-\frac\pi{2}}^{0}{\dfrac{1}{\log^{-1}(t+\frac{\pi}2)}}dt$$ then I substituted: ($y=t+\frac{\pi}2$)
$$\int_{-\frac\pi{2}}^{0}{\dfrac{1}{\log^{-1}(t+\frac{\pi}2)}}dt=\int_{0}^{\frac{\pi}2}{\dfrac{1}{\log^{-1}(y)}}dy$$ which converges (it should converge negatively but I got positively) because $\log^{-1}(y)<\sqrt{y}$ (for example).
The same can be said for $\int_{0}^{\frac\pi{2}}{\log(\cos t)}dt$ so $D_{F(x)}=\left[-\dfrac\pi{2},\dfrac\pi{2}\right]$, am I right?
I have got another problem though: I checked on Desmos and the integral converges negatively where I thought positively.
In order for this function to be well-defined, we need, at first that $\log(\cos(t))$ is well defined. So, we need that: $$\cos(t)>0\Rightarrow t\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$ or every other interval $\left(2k\pi-\frac{\pi}{2},2k\pi+\frac{\pi}{2}\right)$ for any $k\in\mathbb{Z}$. So, $x$ can be, also, any number in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, since $0\leq t\leq x$.
To conclude, $$D_F=\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$
Edit: We could extend $F$ to $\bigcup_{k\in\mathbb{Z}}\left(2k\pi-\frac{\pi}{2},2k\pi+\frac{\pi}{2}\right)$ by setting: $$F(x)=-\int_{A_x}\log(\cos t)dt$$ where $$A_x=(0,x)\cap\left(\bigcup_{k\in\mathbb{Z}}\left(2k\pi-\frac{\pi}{2},2k\pi+\frac{\pi}{2}\right)\right)$$ if $x>0$ and $$A_x=(x,0)\cap\left(\bigcup_{k\in\mathbb{Z}}\left(2k\pi-\frac{\pi}{2},2k\pi+\frac{\pi}{2}\right)\right)$$ if $x<0$.