Calculate $\lim_{n\rightarrow \infty }\int_{0}^{\pi /2}\sqrt[n]{\sin^nx+\cos^nx}\,dx$ I solved an interesting limit sometime, maybe someone will suggest a simpler solution, perhaps through the Lebesgue measure.
\begin{align*}
\lim_{n\rightarrow \infty }\int_{0}^{\pi /2}\sqrt[n]{\sin^nx+\cos^nx}\,dx&=\lim_{n\rightarrow \infty }\left [\int_{0}^{\pi /4}\sqrt[n]{\sin^nx+\cos^nx}\,dx+\int_{\pi /4}^{\pi /2}\sqrt[n]{\sin^nx+\cos^nx} \,dx\right ]\\
&=\lim_{n\rightarrow \infty} \left [ \int_{0}^{\pi /4}\sqrt[n]{\sin^nx+\cos^nx}\,dx+\int_{-\pi /4}^{0}\sqrt[n]{\cos^nx+\left ( -\sin x \right )^n} \,dx\right]\\&
=2\lim_{n\rightarrow \infty }\int_{0}^{\pi /4}\sqrt[n]{\sin^nx+\cos^nx}\,dx
\\
&=2\lim_{n\rightarrow \infty }\int_{0}^{\pi /4}\cos x\sqrt[n]{\tan^nx+1}\,dx\\
&=\sqrt{2}
\end{align*}
$$\int_{0}^{\pi /4}\cos x\,dx<\int_{0}^{\pi /4}\operatorname{cos}x\sqrt[n]{\tan^nx+1}\,dx<\sqrt[n]{2}\int_{0}^{\pi /4}\cos x\,dx$$
 A: If $x\in [0,\pi,2]$, then
$$
\max\{\sin^nx,\cos^nx\}\le \sin^nx+\cos^nx\le 2\max\{\sin^nx,\cos^nx\}
$$
and hence
$$
\max\{\sin x,\cos x\}\le \sqrt[n]{\sin^nx+\cos^nx}\le 2^{1/n}\max\{\sin x,\cos x\}
$$
Now,
$$
\int_0^{\pi/2}\max\{\sin x,\cos x\}\,dx=\int_0^{\pi/4}\cos x\,dx+\int_{\pi/4}^{\pi/2}\sin x\,dx=\sin x\,\big|_0^{\pi/4}-\cos x\,\big|_{\pi/4}^{\pi/2}=\sqrt{2}.
$$
Hence
$$
\sqrt{2}\le\int_0^{\pi/2}\sqrt[n]{\sin^nx+\cos^nx}\,dx\le 2^{1/n}\sqrt{2}
$$
which implies that
$$
\lim_{n\to\infty}\int_0^{\pi/2}\sqrt[n]{\sin^nx+\cos^nx}\,dx=\sqrt{2}.
$$
A: Alternative solution: for any $x\in\left[0,\pi/2\right]$ the quantity $\sqrt[n]{\frac{\sin^n x+\cos^n x}{2}}$ is the mean of order $n$ between $\sin(x)$ and $\cos(x)$, hence $\lim_{n\to +\infty}\sqrt[n]{\frac{\sin^n x+\cos^n x}{2}}=\max(\sin x,\cos x)$. Since $2^{1/n}\to 1$, by the dominated convergence theorem we have
$$ \lim_{n\to +\infty}\int_{0}^{\pi/2}\sqrt[n]{\sin^n x+\cos^n x}\,dx = \int_{0}^{\pi/2}\max(\sin x,\cos x)\,dx = 2\int_{0}^{\pi/4}\cos x\,dx = \sqrt{2}. $$
A: Hint:
For $x<\dfrac\pi4$,
$$\sqrt[n]{\sin^nx+\cos^nx}=\cos x\sqrt[n]{\tan^nx+1}.$$ As $\tan x<1$, the root tends to $1$ and the function is asymptotic to $\cos x$. By symmetry, you get a $\sin x$ behavior past $\dfrac\pi4$.

A: A (not really simpler) alternative to your last line is
$$\begin{align}0<\int_0^{\pi/4} \cos x\left(\sqrt[n]{1 +\tan^n x}-1\right)\,dx&<\int_0^{\pi/4} \cos x\tan^n x\,dx\\
&<\int_0^{\pi/4}\left(1+\tan^2x\right)\tan^n x\,dx\\
&=\frac1{n+1}\to0
\end{align}$$
