Limit of a definite integral We need to calculate $$\lim_{x \to 0}\int_{\sin x}^{x}\frac{dt}{t^3(1+\sqrt{t})}$$
Integral itself doesn't seem to be the problem here. When making a substitution $\sqrt{t}=u$, we get $$\lim_{x \to 0}2\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5(1+u)}=2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5+u^6}$$ 
Then by partial fractions, which I did manually and chcecked with WolframAlpha afterwards, it becomes $$\begin {align} 
2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}\left(\frac{1}{u^5}-\frac{1}{u^4}+\frac{1}{u^3}-\frac{1}{u^2}+\frac{1}{u}-\frac{1}{1+u}\right) du =\\
\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\sqrt{x}}2\left(\log{u}-\log{(1+u)}+\frac{1}{u}-\frac{1}{2u^2}+\frac{1}{3u^3}-\frac{1}{4u^4}\right) du =\\
\lim_{x \to 0}\int_{\sin x}^{x}\left(\log{t}-2\log{(1+\sqrt{t})}+\frac{2}{\sqrt{t}}-\frac{1}{t}+\frac{2}{3t^{3/2}}-\frac{1}{2t^2}\right) dt
\\\end{align}$$
Fianlly we obtain the following limit:
$$\lim_{x \to 0}\left(\log {x}-\log {\sin x}+2\log {(1+\sqrt{x})}-2\log {(1+\sqrt{\sin x})}+\frac{2}{\sqrt{x}}-\frac{2}{\sqrt{\sin x}}-\frac{1}{x}+\frac{1}{\sin x}+\frac{2}{3x^{3/2}}-\frac{2}{3\sin^{3/2} x}-\frac{1}{2x^2}+\frac{1}{\sin^2 x}\right)$$
Here's where I stuck. It gets messy when I try to calculate $\frac{2}{3x^{3/2}}-\frac{2}{3\sin^{3/2} x}$ and $\frac{2}{\sqrt{x}}-\frac{2}{\sqrt{\sin x}}$. The rest is rather doable - de l'Hospital's rule is useful with $-\frac{1}{x}+\frac{1}{\sin x}$ which is $0$ in limit, so as logarithm expressions (obviously) and Taylor expansion helps with $-\frac{1}{2x^2}+\frac{1}{\sin^2 x}$ which, in turn, equals $1/6$ when $x$ approaches $0$.
Did I make any mistakes? I hope not, but even so I'm not certain what to do with this horrible limit. I'd be glad if anyone could point out what to do.

This is not much different from Sami Ben Romdhane's answer.
Since the function $g(t)=\frac{1}{t^3(1+\sqrt{t})}$ is decreasing over $\mathbb{R}^+$, by the mean value theorem:
$$(x-\sin x) \frac{1}{\sin^3 x\,(1+\sqrt{\sin x})}\leq\int_{\sin x}^{x}\frac{dt}{t^3(1+\sqrt{t})}\leq (x-\sin x)\frac{1}{x^3(1+\sqrt{x})},$$
but when $x$ approaches $0$ both the LHS and the RHS approaches $\frac{1}{6}$.
