Limit of integral $\lim_{x\rightarrow 0}\frac{\int_{\cos x}^{1}{}\arctan t ^2\ \mathrm dt}{\tan^2 x}$ $$\lim_{x\rightarrow 0}\frac{\int_{\cos x}^{1}{}\arctan t
^2\ \mathrm dt}{\tan^2 x}$$
There is a part of the answer written as
$$\lim_{x\rightarrow 0}\int_{\cos x}^{1}{}\arctan t
^2\ \mathrm dt=\lim_{x\rightarrow 0}\tan^2 x=0$$
How do you even come to this conclusion?
 A: Note that for $x\in (-\pi/4,\pi/4)$, we have
$$(1-\cos(x))\arctan(\cos^2(x))\le \int_{\cos(x)}^1 \arctan(t^2)\,dt\le (1-\cos(x))\frac\pi4$$
Hence, we assert that 
$$ \frac{1-\cos(x)}{\tan^2(x)}\arctan(\cos^2(x))\le \frac{\int_{\cos(x)}^1 \arctan(t^2)\,dt}{\tan^2(x)}\le \frac\pi4 \frac{1-\cos(x)}{\tan^2(x)}$$
whence using $\frac{1-\cos(x)}{\tan^2(x)}=\frac{\cos^2(x)}{1+\cos(x)}$ and applying of the squeeze theorem yields the coveted limit
$$\lim_{x\to 0}\frac{\int_{\cos(x)}^1 \arctan(t^2)\,dt}{\tan^2(x)}=\frac\pi8$$
A: To answer your question: They first check whether L'Hospital rule is applicable.
Since $\lim_{x\to 0} \tan^2 x = 0$ they need to check first whether 
$\lim_{x\rightarrow 0}\int_{\cos x}^{1}\arctan t
^2\mathrm dt=0$, as well.
This is so, since $\arctan t^2$ is continuous and, hence, you have $0\leq \arctan t^2 \leq M$ for $t\in\left[\frac 12,1\right]$, for example. So, for $0 \leq x\leq \frac{\pi}{3}$  (note that $\cos \frac{\pi}{3} = \frac 12$) you have $\cos x \in \left[\frac 12,1\right]$ and hence
$$0\leq \int_{\cos x}^{1}\arctan t
^2\mathrm dt\leq M(1-\cos x) \stackrel{x\to 0}{\longrightarrow}0$$
Now, L'Hospital gives
\begin{eqnarray}\frac{\int_{\cos x}^{1}{}\arctan t
^2\mathrm dt}{\tan^2 x} 
& \stackrel{L'Hosp.}{\sim} & \frac{-\arctan \cos^2 x \cdot (-\sin x)}{2\frac{\sin x}{\cos^3 x}} \\& = & \frac 12 \cos^3 x\arctan \cos^2 x\\
& \stackrel{x\to 0}{\longrightarrow} &\frac 12 \frac{\pi}{4} = \frac{\pi}{8}
\end{eqnarray}
A: You can use L'Hospital and the fundamental theorem of calculus since
$$\frac d {dx} \int_{a(x)}^1 f(t)\,dt=-f(a(x))\, a'(x)$$ So, for your case, using  L'Hospital once, we arrive at $$\lim_{x\rightarrow 0}\frac{\int_{\cos (x)}^{1}{}\arctan (t
^2)\mathrm dt}{\tan^2 x}=\lim_{x\rightarrow 0}\left(\frac{1}{2} \cos ^3(x) \tan ^{-1}\left(\cos ^2(x)\right)\right)=\frac \pi 8$$
Added for your curiosity.
It happens that the antiderivative has a closed form expression (try Wolfram Alpha for it).
Using the bounds and Taylor expansions, we should have as a result
$$\frac{\pi }{8}-\left(\frac{1}{8}+\frac{3 \pi }{32}\right)
   x^2+\left(\frac{1}{12}+\frac{\pi }{64}\right) x^4+O\left(x^6\right)$$
