Find $\lim_{n\to \infty}\int _0^{\frac{\pi}{2}} \sqrt{1+\sin^nx}$ 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.
 A: Let's define the following sequence:
$$J_n=\int _0^{\frac{\pi}{2}} \sin^nx\, dx$$
Since $\sin^n x \geq 0$, for $x \in [0,\frac{\pi}{2}]$, we have:
$$|\sqrt{1+\sin^n x}-1|=\frac{\sin^n x}{\sqrt{1+\sin^n x}+1} \leq \frac{\sin^n x}{2}$$
Therefore
$$1-\frac{1}{2}\sin^n x\leq \sqrt{1+\sin^n x}\leq 1+\frac{1}{2}\sin^n x$$
and integrating:
$$\frac{\pi}{2}-\frac{1}{2}J_n \leq I_n \leq \frac{\pi}{2}+\frac{1}{2}J_n\ \ \ \ \ \ \ \ (*)$$
Now, integrating by parts, we can deduce that:
$$J_n=\frac{n-1}{n}J_{n-2}\Rightarrow nJ_nJ_{n-1}=(n-1)J_{n-1}J_{n-2}$$
Therefore
$$nJ_nJ_{n-1}=(n-1)J_{n-1}J_{n-2}=(n-2)J_{n-2}J_{n-3}=...=J_1J_0=\frac{\pi}{2}$$
Clearly $J_n$ is convergent, and its limit must be $0$. Therefore, squeezing in $(*)$:
$$\lim_{n\to \infty}I_n = \frac{\pi}{2}$$
A: Hint: Use Dominated convergence theorem.
A: DCT is the way to go if you have that at your disposal. Here's an elementary solution: Let $0<b<\pi/2.$ Then the $n$th integral, let's call it $I_n,$ satisfies
$$\pi/2 = \int_0^{\pi/2}1\,dx < I_n < b\cdot \sqrt {1+\sin^n b} + \sqrt 2(\pi/2-b).$$
Taking limits, we get
$$\pi/2 \le \lim_{n\to \infty} I_n \le b\cdot 1 + \sqrt 2(\pi/2-b).$$
Now let $b\to \pi/2^-$ to see $\pi/2 \le \lim I_n \le \pi/2.$
A: 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 $$n\ge n_{\delta}\implies \pi/2-2\delta <\int_0^{\pi/2}f_n(x)dx <\pi/2+2\delta.$$
