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.
 A: Your proof is nice but there's more simple:
Let consider this simplified integral
$$\int_{\sin x}^x\frac{dt}{t^3}=-\frac12t^{-2}\Bigg|_{\sin x}^x=-\frac12\left(\frac1{x^2}-\frac1{\sin^2x}\right)\sim_0-\frac12\frac{(x-\frac{x^3}{6})^2-x^2}{x^4}\sim_0\frac16$$
and now we prove that the two integrals have the same limit by:
\begin{align}
0\le\int_{\sin x}^x\left(\frac{1}{t^{3}}-\frac{1}{t^{3}(1+\sqrt t)}\right)dt=\int_{\sin x}^x\frac{\sqrt t}{t^3(1+\sqrt t)}dt &\le\int_{\sin x}^x t^{-5/2}dt = -\frac25t^{-3/2}\Bigg|_{\sin x}^x \\ 
&=-\frac25\left(\frac1{x^{3/2}}-\frac1{\sin^{3/2}x}\right)\sim_0\frac25\frac{x^{7/2}}{4x^3}\xrightarrow{x\to0}0
\end{align}
A: 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 $\frac{1}{6}$.
A: When you substitute a new variable in, you must not forget to change the limits of your integral.
So,
instead of $$x, \sin x$$
We have 
$$\sqrt{x}, \sqrt{\sin x}$$
Everything else looks correct.
EDIT
To solve the limit:
$$\lim_{x\to 0} \frac{2}{\sqrt{x}} + \frac{2}{\sqrt{\sin x}}$$
$$ = \lim_{x\to 0} \frac{2(\sqrt{\sin{x}} - \sqrt{x})}{\sqrt{x}\sqrt{\sin x}}$$
Use l'hopsitals
Can you get it from here?
