TITLE: Find $\lim_{n\to \infty}\int _0^{\frac{\pi}{2}} \sqrt{1+\sin^nx}$
QUESTION [6 upvotes]: Define
$$I_n=\int _0^{\frac{\pi}{2}} \sqrt{1+\sin^nx}\, dx$$
I have to show this sequence is convergent and find its limit. I proved it is decreasing: $\sin^{n+1} x \le \sin^n x \implies I_{n+1} \le I_n$. Also, it is bounded because:
$$0 \le \sin^n x \le 1 \implies \frac\pi{2} \le \int _0^{\frac{\pi}{2}} \sqrt{1+\sin^n x}\, dx\le \frac{\pi\sqrt{2}}{2}$$
so it is convergent. I'm stuck at finding the limit. I think it should $\frac{\pi}{2}$ but I'm not sure.

REPLY [2 votes]: Take any $\delta \in (0,\pi/2).$ The sequence of integrable functions $f_n(x)=\sqrt {1+\sin^n x}$ converges uniformly to $1$ on $[0,\pi/2-\delta].$ So $\int_0^{\pi/2-\delta} f_n(x)dx$ converges to $\int_0^{\pi/2-\delta} 1\cdot dx=\pi/2 -\delta.$ So there exists $n_{\delta}\in \Bbb N$ such that $$n\ge n_{\delta}\implies  -\delta<\int_0^{\pi/2-\delta}f_n(x)dx-(\pi/2-\delta)<\delta.$$ Meanwhile, for all $n$ we have $$0<\int_{\pi/2-\delta}^{\pi/2}f_n(x)dx<\int_{\pi/2-\delta}^{\pi/2}2\cdot dx=2\delta.$$ Therefore $$n\ge n_{\delta}\implies \pi/2-2\delta <\int_0^{\pi/2}f_n(x)dx <\pi/2+2\delta.$$