Compute $F(x)$ and $\displaystyle{\lim_{x \to \frac{\pi}{2}}} F(x)$

Question

Consider the function F defined on $]-\frac{\pi}{2}; \frac{\pi}{2}[$ by

$$F(x) = \int_0^x t \tan^2{t} dt$$

1. Compute $F(x)$ and $\displaystyle{\lim_{x \to \frac{\pi}{2}}} F(x)$
2. Determine the sign of $F(x)$

My attempt:

\begin{align}F(x) &= \int_0^x t \tan^2{t} dt \\ &= [t \tan{t}-t^2]_0^x - \int_0^x \tan{t} dt + \int_0^x t dt \\ &= x \tan{x}-x^2 + \ln{|\cos{x}|} + \frac{x^2}{2} \\ &= x \tan{x} + \ln{|\cos{x}|} - \frac{x^2}{2}\end{align}

I don't see how to compute the limit knowing that $\displaystyle{\lim_{x \to \frac{\pi}{2}}}\tan{x}$ does not exist.

EDIT: \begin{align}\displaystyle{\lim_{x \to \frac{\pi}{2}}} F(x) &= \displaystyle{\lim_{x \to \frac{\pi}{2}^-}} F(x) \\ &= \displaystyle{\lim_{x \to \frac{\pi}{2}^-}} (x \tan{x} + \ln{|\cos{x}|}) - \displaystyle{\lim_{x \to \frac{\pi}{2}^-}} \frac{x^2}{2} \\ &= \displaystyle{\lim_{x \to \frac{\pi}{2}^-}} \tan{x}(x + \frac{\ln{|\cos{x}|}}{\tan{x}}) - \displaystyle{\lim_{x \to \frac{\pi}{2}^-}} \frac{x^2}{2} \end{align}

and $\displaystyle{\lim_{x \to \frac{\pi}{2}^-}} \frac{\ln{|\cos{x}|}}{\tan{x}} = \displaystyle{\lim_{x \to \frac{\pi}{2}^-}} \frac{\ln{\cos{x}}}{\tan{x}} = 0$

Thus \begin{align}\displaystyle{\lim_{x \to \frac{\pi}{2}}} F(x) &= + \infty\end{align}

Is it like this?

Answer 1

With respect to definition of function in $]-\frac{\pi}{2}; \frac{\pi}{2}[ \ $ , $ \ x $ tends to $\frac{\pi}{2}^{-}$ (from down to $\frac{\pi}{2}$) so $$\lim_{x \to \frac{\pi}{2}^{-}}\tan x=+\infty$$ or $$\lim_{x \to \frac{\pi}{2}^{-}}\tan x=\lim_{x \to \frac{\pi}{2}^{-}}\frac {\sin x}{\cos x}=\frac {1}{0^+}\to +\infty$$