Limit of $\frac{1}{x^4}\int_{\sin x}^{x} \arctan t dt$ I am trying to find this limit,

$$\lim_{x \rightarrow 0} \frac{1}{x^4} \int_{\sin{x}}^{x} \arctan{t}dt$$

Using the fundamental theorem of calculus, part 1,
$\arctan$ is a continuous function, so
$$F(x):=\int_0^x \arctan{t}dt$$
and I can change the limit to
$$\lim_{x \rightarrow 0} \frac{F(x)-F(\sin x)}{x^4}$$
I keep getting $+\infty$, but when I actually integrate $\arctan$ (integration by parts) and plot the function inside the limit, the graph tends to $-\infty$ as $x \rightarrow 0+$.
I tried using l'Hospital's rule, but the calculation gets tedious.
Can anyone give me hints?
EDIT
I kept thinking about the problem, and I thought of power series and solved it, returned to the site and found 3 great answers. Thank You!
 A: Power series can help, if you know them.
We know that
$$
\arctan t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n+1}}{2n+1},\qquad \lvert t\rvert<1.
$$
Therefore an antiderivative for $\arctan t$ is
$$
F(t):=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n+2}}{(2n+1)(2n+2)},\qquad \lvert t\rvert<1.
$$
So, as $x\to0$, 
$$
F(x)=\frac{x^2}{2}-\frac{x^4}{12}+O(x^6).
$$
Now, recalling that
$$
\sin x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{6}+O(x^5)=x+O(x^3),
$$
we see that as $x\to0$ (and therefore $\sin x\to0$),
$$
\begin{align*}
F(\sin x)&=\frac{1}{2}\left(x-\frac{x^3}{6}+O(x^5)\right)^2-\frac{1}{12}\left(x+O(x^3)\right)^4+O[(x+O(x^3))^6]\\
&=\frac{1}{2}\left[x^2-\frac{x^4}{3}+O(x^6)\right]-\frac{1}{12}\left[x^4+O(x^6)\right]+O(x^6)\\
&=\frac{x^2}{2}-\frac{x^4}{4}+O(x^6)
\end{align*}
$$
So, all told, our limit is
$$
\lim_{x\to0}\frac{F(x)-F(\sin x)}{x^4}=\lim_{x\to0}\frac{\,\frac{x^4}{6}\,}{x^4}=\frac{1}{6}
$$
A: Here's one way to do this. The Taylor expansion for $\arctan(x) = x - x^3/3 + x^5/5 + \cdots$. So then
$$ \int_{\sin(x)}^x \arctan(t)dt \sim (x^2/2 - x^4/12 + x^6/30) - \frac{(\sin x)^2}{2} + \frac{(\sin x)^4}{12} - \frac{(\sin x)^6}{30}.$$
Using another Taylor expansion, $\sin x = x - x^3/3! + x^5/5! + \cdots$, so the plan is to pay attention only to those terms of degree up to $4$ in these expansions. Then (dropping all terms of degree greater than $4$),
$$ (\sin x)^2/2 = (x - x^3/3!)^2/2 = x^2/2 - x^4/6$$
and
$$ (\sin x)^4/12 = (x - x^3/3!)^4/4 = x^4/12.$$
Thus
$$ \int_{\sin(x)}^x \arctan(t)dt \sim (x^2/2 - x^4/12) - (x^2/2 - x^4/6) + x^4/12 = x^4/6.$$
So multiplying by $x^{-4}$ and taking the limit gives $1/6$.
As an aside, l'Hopital's rule would also work.
A: Since $F(0) = 0$ and everything is smooth, you can apply de l'Hopital and get
$$\lim_{x \to 0}\frac{F(x)-F(\sin x)}{x^4} = \lim_{x \to 0}\frac{\arctan x - \cos x\arctan \sin x}{4x^3}.$$
This last limit can be evaluated using Taylor series:
$$\arctan x = x-\frac{x^3}{3}+O(x^5) $$
and
$$\cos x \arctan \sin x = x-x^3+O(x^5) $$
and the limit you are looking for is equal to
$$\lim_{x \to 0}\frac{x-\dfrac{x^3}{3}-x+x^3 + O(x^5)}{4x^3} = \frac{1}{6}. $$
The ''ugly'' Taylor expansion is obtained combining the Taylor expansions of $\sin$, $\cos$ and $\arctan$. Easier done than said.
A: Note that by the Mean Value Theorem for integrals we have $$\int_{\sin x}^{x}\arctan t\,dt = (x - \sin x)\arctan c$$ for some $c$ between $x$ and $\sin x$. Then we have $$\lim_{x \to 0}\frac{1}{x^{4}}\int_{\sin x}^{x}\arctan t\,dt = \lim_{x \to 0}\frac{x - \sin x}{x^{3}}\cdot\frac{\arctan c}{x}$$ The first factor on the right tends to $1/6$ (via L'Hospital's Rule or Taylor series) and we show that next factor tends to $1$. For this we assume that $x \to 0^{+}$ so that $\sin x < c < x$ and therefore $$\arctan \sin x < \arctan c < \arctan x$$ and dividing by $x$ we get $$\frac{\arctan \sin x}{\sin x}\cdot\frac{\sin x}{x} < \frac{\arctan c}{x} < \frac{\arctan x}{x}$$ By Squeeze theorem we see that $(\arctan c)/x \to 1$ as $x \to 0^{+}$. A similar argument can be given for $x \to 0^{-}$ and we have the desired limit as $1/6$.
