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.
 A: 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
$${\pi \over 2}\ln(2)$$
A: 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
$n\int_0^{\frac \pi 2}1-\sqrt[n]{\sin x}dx=\int_0^{\frac \pi 2} -\log \sin x dx +O(\frac  1 n)=\dfrac{\pi \log 2}{2} +O(\frac 1 n)\to\dfrac{\pi \log 2}{2}$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\lim_{n \to \infty}\braces{%
     n\int_{0}^{\pi/2}\,
     \bracks{1 - \root[n]{\,\sin\pars{x}}}\,{\rm d}x}:\ {\large ?}}$

\begin{align}
\int_{0}^{\pi/2}\root[n]{\sin\pars{x}}\,{\rm d}x&=\int_{0}^{1}t^{1/n}
\,{\dd t \over \root{1 - t^{2}}}
=\int_{0}^{1}t^{1/\pars{2n}}\pars{1 - t}^{-1/2}\,\half\,t^{-1/2}\,\dd t
\\[3mm]&=\half\int_{0}^{1}t^{1/\pars{2n} - 1/2}\pars{1 - t}^{-1/2}\,\dd t
=\half\,{\rm B}\pars{{1 \over 2n} + \half,\half}
\\[3mm]&=\half\,{\Gamma\pars{1/\bracks{2n} + 1/2}\Gamma\pars{1/2} \over \Gamma\pars{1/\bracks{2n} + 1}}
\end{align}

When $\ds{n \gg 1}$:
\begin{align}
\int_{0}^{\pi/2}\root[n]{\sin\pars{x}}\,{\rm d}x
&\approx
{\root{\pi} \over 2}\,{\Gamma\pars{1/2} + \Gamma\pars{1/2}\Psi\pars{1/2}/\pars{2n}
 \over \Gamma\pars{1} + \Gamma\pars{1}\Psi\pars{1}/\pars{2n}}
={\pi \over 2}\,{1 + \Psi\pars{1/2}/\pars{2n} \over 1 + \Psi\pars{1}/\pars{2n}}
\\[3mm]&\approx
{\pi \over 2}\,\bracks{1 + {\Psi\pars{1/2} \over 2n}}
\bracks{1 - {\Psi\pars{1} \over 2n}}
\approx {\pi \over 2}\,\bracks{1 + {\Psi\pars{1/2} - \Psi\pars{1} \over 2n}}
\end{align}

$$
\color{#00f}{%
\lim_{n \to \infty}\braces{n\int_{0}^{\pi/2}\,\bracks{1 - \root[n]{\,\sin\pars{x}}}
\,{\rm d}x}}
={\pi \over 4}\,\bracks{\Psi\pars{1} - \Psi\pars{\half}}
=\color{#00f}{\half\,\pi\ln\pars{2}}
$$
  since $\ds{\Psi\pars{1} = -\gamma}$ and
  $\ds{\Psi\pars{\half} = -\gamma - 2\ln\pars{2}}$. See
  this table.

A: 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
$$\lim_{n \to \infty}\int_{0}^{\pi/2} n(1 - \sqrt[n]{\sin x})\text{d}x = \int_{0}^{\pi/2} \lim_{n \to \infty}n(1 - \sqrt[n]{\sin x}) \text{d}x= \int_{0}^{\pi/2} -\log \sin x \text{d}x = \frac{\pi \log 2}{2}$$
A: You can have a close form solution; infact if $Re(1/n)>-1$ you have that the integral collapse in:
$$\int_{0}^{\pi/2}\left[1-(\sin(x))^{1/n}\right]dx=\frac{1}{2} \left(\pi -\frac{2 \sqrt{\pi } n \Gamma
   \left(\frac{n+1}{2 n}\right)}{\Gamma
   \left(\frac{1}{2 n}\right)}\right)$$
So we define:
$$y(n)=\frac{n}{2} \left(\pi -\frac{2 \sqrt{\pi } n \Gamma
   \left(\frac{n+1}{2 n}\right)}{\Gamma
   \left(\frac{1}{2 n}\right)}\right)$$
And performing the limit:
$$\lim_{n \rightarrow + \infty}y(n)=-\frac{1}{4} \pi  \left[\gamma +\psi
   ^{(0)}\left(\frac{1}{2}\right)\right]$$
A: 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.
