Evaluating $\lim \limits_{n\to \infty}\,\,\, n\!\! \int\limits_{0}^{\pi/2}\!\! \left(1-\sqrt [n]{\sin x} \right)\,\mathrm dx$

You can use the following fact

$$f(x) = \log \sin x$$ is integrable in $(0, \pi/2)$

and

$$\int_{0}^{\pi/2} -\log \sin x \text{d}x = \frac{\pi \log 2}{2}$$

Now by the mean value theorem (applied to $(\sin x)^y$, as a function of $y$), we have that

for some $c \in (0, \frac{1}{n})$

$$ \dfrac{1 - \sqrt[n]{\sin x}}{\frac{1}{n}} = -(\sin x)^c \log \sin x \le -\log \sin x$$

Since $\log \sin x$ is integrable, by the dominated convergence theorem, we can take the limit inside the integral to get

$$\lim_{n \to \infty}\int_{0}^{\pi/2} n(1 - \sqrt[n]{\sin x})\text{d}x = \int_{0}^{\pi/2} \lim_{n \to \infty}n(1 - \sqrt[n]{\sin x}) \text{d}x= \int_{0}^{\pi/2} -\log \sin x \text{d}x = \frac{\pi \log 2}{2}$$

This is equivalent to finding $\lim_{\epsilon \rightarrow 0} {f(\epsilon) \over \epsilon}$, where $f(\epsilon) = \int_0^{\pi \over 2} (1 - \sin(x)^{\epsilon})\,dx$. Since $\lim_{\epsilon \rightarrow 0} \sin(x)^{\epsilon} = 1$, one has $\lim_{\epsilon \rightarrow 0} f(\epsilon) = 0$, and so by L'Hopital's rule you get $$\lim_{\epsilon \rightarrow 0} {f(\epsilon) \over \epsilon} = \lim_{\epsilon \rightarrow 0} f'(\epsilon)$$ Differentiating under the integral sign gives $$f'(\epsilon) = -\int_0^{\pi \over 2} \ln(\sin(x))(\sin(x))^{\epsilon}\,dx$$ The limit of this as $\epsilon \rightarrow 0$ is $$-\int_0^{\pi \over 2} \ln(\sin(x))\,dx$$ This integral is well-known (and I'm sure it's been done on this site), and the above is just $${\pi \over 2}\ln(2)$$

This is only a different way to get it to the final integral , but here goes, $\sqrt[n]{\sin x}=\exp(\dfrac{\log \sin x}{n})=1+\dfrac{\log \sin x}{n}+ o(\frac{1}{n})$

So, $1-\sqrt[n]{\sin x}=-\dfrac{\log \sin x}{n} +o(\frac 1 n)$

Using, this we get $n\int_0^{\frac \pi 2}1-\sqrt[n]{\sin x}dx=\int_0^{\frac \pi 2} -\log \sin x dx +O(\frac 1 n)=\dfrac{\pi \log 2}{2} +O(\frac 1 n)\to\dfrac{\pi \log 2}{2}$