Compute $\lim_{n\to\infty}\int_0^\frac{\pi}{2} \frac{\sin^nx}{\sqrt{1+x}}\, dx$ The question is the one from the title i.e. :
Compute $\displaystyle\lim_\limits{{n\to\infty}}\int_0^\frac{\pi}{2} \frac{\sin^nx}
 {\sqrt{1+x}}\, dx$
I do realise that I should probably use Monotone Convergence Theorem or Dominated Convergence Theorem. 
The problem is that:


*

*To use the first one I shoul prove first that $\frac{\sin^nx}{\sqrt{1+x}} \leq \frac{\sin^{n+1}x}{\sqrt{1+x}}$ for all $x\in(0,\frac{\pi}{2})$ which I think is not true. I would rather claim that the opposite is true but I am struggling with proving it, I mean : $\frac{\sin^nx}{\sqrt{1+x}} \geq \frac{\sin^{n+1}x}{\sqrt{1+x}}$ for all $x\in(0,\frac{\pi}{2})$. 

*To use the second theorem I need $\frac{\sin^nx}{\sqrt{1+x}}$ to be bbd by some measurable function I can take $1$ so this bit is easy but also I need pointwise convergence to some function through all the $(0,\frac{\pi}{2})$  and here the funny part starts. Using some numerical (not sophisticated ;) ) methods ( by saying that I mean I just checked the behaviour for large n ;) ) It looks like $0$ (- zero) is a good candidate, but I couldn't prove it.


So the question is: 


*

*How to prove the second inequality in 1.) 


Or


*

*How to prove that $\lim_\limits{{n\to\infty}}\frac{\sin^nx}{\sqrt{1+x}}=0$ on $(0,\frac{\pi}{2})$

 A: The given limit is clearly zero.
$$\begin{eqnarray*} 0\leq \int_{0}^{\pi/2}\frac{\sin^n(x)}{\sqrt{1+x}}\,dx &\color{red}{\leq}& \int_{0}^{\pi/2}\sin^n(x)\,dx =\int_{0}^{\pi/2}\cos^n(x)\,dx \leq \int_{0}^{\pi/2}e^{-nx^2/2}\,dx\\ &\leq& \int_{0}^{+\infty} e^{-nx^2/2}\,dx = \color{red}{\sqrt{\frac{\pi}{2n}}}\end{eqnarray*}$$
As an alternative, the given integral is $\leq \int_{0}^{1}\frac{x^n}{\sqrt{1-x^2}}$. Over the interval $(0,1)$ the sequence of functions $f_n(x)=x^n$ is pointwise convergent to $0$ and bounded by $1$. Since $\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{2}$, the claim follows from the dominated convergence theorem, too.
A: As Jack D'Aurizio answered $$ 0\leq \int_{0}^{\pi/2}\frac{\sin^n(x)}{\sqrt{1+x}}\,dx\, {\leq} \int_{0}^{\pi/2}\sin^n(x)\,dx =\frac{\sqrt{\pi }} 2\frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma
   \left(\frac{n}{2}+1\right)}$$ Now, using Stirling approximation $$\frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma
   \left(\frac{n}{2}+1\right)}=\frac 1 {\sqrt{ n}}\left(\sqrt{2}-\frac{1}{2 \sqrt{2} n}+O\left(\frac{1}{n^2}\right)\right) $$
A: The integrands converge pointwise to $0$ on $[0,\pi/2).$ By the dominated convergence theorem (the dominating function being $1/\sqrt {1+x}$), the limit is $0.$
For a more elementary solutionn, let $0<h<\pi/2.$ Then
$$ 0\le \int_0^{\pi/2} \frac{\sin^n x}{\sqrt {1+x}}\,dx \le \int_0^{\pi/2} \sin^n x\,dx$$ $$\tag 1 = \int_0^{\pi/2-h} \sin^n x\,dx + \int_{\pi/2-h}^{\pi/2} \sin^n x\,dx \le \sin^n(\pi/2 -h)\cdot (\pi/2) + 1\cdot h.$$
Because $0\le \sin(\pi/2 -h) <1,$ $\sin^n(\pi/2 -h) \to 0.$ It follows that the $\limsup$ of the left side of $(1)$ is $\le h.$ Because $h$ is arbitrarily small, this $\limsup$ is $0.$ Thus the desired limit is $0.$ 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\lim_{n \to \infty}\int_{0}^{\pi/2}{\sin^{n}\pars{x} \over \root{1 + x}}\,\dd x & =
\lim_{n \to \infty}\int_{0}^{\pi/2}
{\cos^{n}\pars{x} \over \root{1 + \pars{\pi/2 - x}}}\,\dd x =
{1 \over \root{1 + \pi/2}}\lim_{n \to \infty}\int_{0}^{\infty}
\expo{-nx^{2}/2}\,\dd x
\\[5mm] & =
{1 \over \root{1 + \pi/2}}\lim_{n \to \infty}\root{\pi \over 2n} = \bbx{0}
\end{align}
