Calculate $\lim_{n\rightarrow \infty }\int_{0}^{\pi /2}\sqrt[n]{\sin^nx+\cos^nx}\,dx$

Question

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*}
$$\int_{0}^{\pi /4}\cos x\,dx<\int_{0}^{\pi /4}\operatorname{cos}x\sqrt[n]{\tan^nx+1}\,dx<\sqrt[n]{2}\int_{0}^{\pi /4}\cos x\,dx$$

Answer 1

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 $$ \lim_{n\to\infty}\int_0^{\pi/2}\sqrt[n]{\sin^nx+\cos^nx}\,dx=\sqrt{2}. $$