Limit of integral with sinx I need to find this limit:

$$\lim\limits_{x\rightarrow 0}\int_{\sin x}^{x}\dfrac{1}{t^3(1+t^{1/2})}\,dt.$$

Thank you for any help.
$\int_{\sin x}^x\frac{1}{t^3(1+t^{1/2})}dt&gt;\int_{\sin x}^x\frac{1}{x^3(1+x^{1/2})}dt=(x-\sin x)\frac{1}{x^3(1+x^{1/2})}$. Then using the taylor series of $\sin x$, you get the lower bound. The upper bound is in a similar way. Actually, you do not need to solve the integral.
Substitute $x^2=t$ and write the inner integral as $$2\int {1+x-x-x^2+x^2+x^3-x^3-x^4+x^4+x^5-x^5\over x^5(1+x)}$$$$=2\left(\int{1\over x^5}dx-\int{1\over x^4}dx+\int{1\over x^3}dx-\int{1\over x^2}dx+\int{1\over x}dx-\int{1\over 1+x}dx\right)$$

Now integrate and put the limits.See what you get..!