how to find $\lim\limits_{x\to 0} \int_0^x \frac{\sin^2(u)}{\sin(x^3)}du $ How to find
$$\lim\limits_{x\to 0} \int_0^x \frac{\sin^2(u)}{\sin(x^3)}du ?
$$
Where should I even start with this one? Can I pull out $\frac{1}{\sin(x^3)}$ outside the integral as $x$ is an constant and we're finding integral with respect to to $u$?, and then apply Leibniz integral rule?

Where should I even get started with this one its kinda confusing.


$$\lim_{x \to 0} \left(\frac{1}{\sin(x^3)} \int_0^x \sin^2 u \ du \right)$$
$$= \lim_{x \to 0} \left(  \frac{1}{\sin(x^3)} \left[\frac{1}{2}u - \frac{1}{4} \sin2u \right]_0^x \right)$$
and when $x^3$ is small, $\sin x \approx x - \frac{x^3}{3!}$ and $\sin x^3 \approx x^3$, so:

 $$= \lim_{x \to 0} \left(\frac{1}{x^3} \left[\frac{1}{2}x - \frac{1}{4} \sin(2x) \right] \right)$$
$$= \lim_{x \to 0} \left(\frac{1}{x^3} \left[\frac{1}{2}x - \frac{1}{4} (2x - 8x^3/6) \right] \right)$$
$$= \lim_{x \to 0} \left(\frac{1}{x^3} (2x^3/6)  \right) =\frac{1}{3}.$$

