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

I have done the problem .
My method:
First I applied L'Hôpital's rule as it can be made of the form $\frac0 0$. Then I used weighted mean value theorem and using sandwich theorem reduced the limit to an integral which could be evaluated using properties of define integration .
I would like to see other different ways to solve for the limit.

Result
Let
$$f(n) = n \int_0^{\frac{\pi}{2}} (1- \sin(x)^{\frac{1}{n}})\, dx$$
then
$$\lim_{n\to \infty } \, f( n)=\frac{\pi}{2} \log(2) \simeq 1.088793045151801$$
Derivation 1
Attempting to perform the limit directly under the integral we write
$$n(1-\sin(x)^\frac{1}{n}) = n\left(1-\exp\left(\frac{1}{n} \log\left(\sin(x)\right)\right)\right)\\ 
\simeq  n\left(1-1 - (\frac{1}{n}) \log(\sin(x))+O(\frac{1}{n^2})\right) \\
= -  \log(\sin(x)) + O(\frac{1}{n})$$
and the limiting integral becomes
$$- \int_0^\frac{\pi}{2}  \log(\sin(x)) = \frac{\pi}{2} \log(2)$$
Derivation 2
The transformation $\sin(x) \to t$, $dx \to \frac{dt}{\sqrt{1-t^2}}$leads to
$$f(n)= \int_0^1  n \left(1-t^{1/n}\right) \frac{1}{\sqrt{1-t^2}} \, dt$$
Now performing the limit under the integral sign gives
$$\lim_{n\to \infty } \, n \left(1-t^{1/n}\right)=-\log (t)$$
so that the integral becomes
$$- \int_0^1 \frac{\log (t)}{\sqrt{1-t^2}} \, dt=\frac{\pi}{2} \log(2)$$
which is the announced result.
