Compute $\lim_{n\rightarrow\infty}\int_{0}^{\frac{\pi}{2}}\frac{n\cos x \sin^n x}{1+x}dx$ I'm trying to compute $\lim_{n\rightarrow\infty}\int_{0}^{\frac{\pi}{2}}\frac{n\cos x \sin^n x}{1+x}dx$. 
In the given interval $(0,\frac{\pi}{2})$, $\frac{n\cos x \sin^n x}{1+x}$ is non negative, and increasing, I used the monotone convergence theorem to interchange the limit and the integral to get $$\lim_{n\rightarrow\infty}\int_{0}^{\frac{\pi}{2}}\frac{n\cos x \sin^n x}{1+x}dx=\int_{0}^{\frac{\pi}{2}}\lim_{n\rightarrow\infty}\frac{n\cos x \sin^n x}{1+x}dx$$ 
But when I'm trying to compute the limit, I'm stuck. I even tried to do through the "$u$ substitution" as well but the denominator isn't helping. Any input is appreciated.
 A: Here's a possible approach. We have
$$
\int_0^{\pi/2} \frac{\cos x \sin^n x}{1+x } dx = \int_0^{\pi/2} \frac{\sin^n x}{1+x } d \sin x = (\text{set } y = \sin x) = \int_0^1 \frac{y^n}{ 1 + \arcsin y} dy.
$$
Now, the last integral does not seem a very pleasant one to compute explicitly, but we can compute the desired limit. Just observe that the main contribution comes from the neighborhood of 1. 
Namely, fix any $0<\alpha <1$, then 
$$
n \int_0^\alpha \frac{y^n}{ 1 + \arcsin y}  dy \leq n \int_0^\alpha y^n dy  =\frac{n}{n+1} \alpha^n \to 0 ,  \text{ as } n \to \infty,
$$ 
as $\alpha < 1$. Hence, the contribution comes from $\int_\alpha^1$, where we have
$$
\frac{y^n}{1 + \pi/2} \leq \frac{y^n}{1 + \arcsin y} \leq \frac{y^n}{1 + \arcsin \alpha}.
$$
Now integrating the last ineqalities over $[\alpha, 1]$ we obtain
$$
\frac{n}{n+1} \frac{1 - \alpha^{n+1}}{1+\pi/2} \leq n \int_{\alpha}^1 \frac{y^n}{1 + \arcsin y} dy\leq \frac{n}{n+1} \frac{1 - \alpha^{n+1}}{1 + \arcsin \alpha}.
$$
Finally, taking $n\to \infty$ in the last line,  we get
$$
 \frac{1}{1+\pi/2} \leq \lim n \int_{\alpha}^1 \frac{y^n}{1 + \arcsin y} dy\leq  \frac{1 }{1 + \arcsin \alpha}.
$$
But we also proved that the limit of the integral over $[0, \alpha]$ is vanishing, hence the limit in the last inequality is the actual limit in question. But as $\alpha<1$ is arbitrary, we get that the limit has to be equal to 
$$
\frac{1}{1+\pi/2}.
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\lim_{n \to \infty}\int_{0}^{\pi/2}
{n\cos\pars{x}\sin^{n}\pars{x} \over 1 + x}\,\dd x & =
\lim_{n \to \infty}\bracks{{n \over n + 1}\int_{x\ =\ 0}^{x\ =\ \pi/2}
{\dd\sin^{n + 1}\pars{x} \over 1 + x}}
\\[5mm] & =
\lim_{n \to \infty}\bracks{{n \over n + 1}\,\color{red}{1 \over 1 + \pi/2} +
{n \over n + 1}\int_{0}^{\pi/2}
{\cos^{n + 1}\pars{x} \over \pars{1 + \pi/2 - x}^{2}}\,\dd x}
\\[5mm] & = \bbx{\large{1 \over 1 + \pi/2}} \approx 0.3890
\end{align}

$$
\overbrace{\int_{0}^{\pi/2}
{\cos^{n + 1}\pars{x} \over \pars{1 + \pi/2 - x}^{2}}\,\dd x
\sim {1 \over \pars{1 + \pi/2}^{\, 2}}\
\underbrace{\int_{0}^{\infty}{\expo{-\pars{n + 1}x^{2}/2}}\dd x}
_{\ds{\root{\pi \over 2}{1 \over \pars{n + 1}^{1/2}}}} \color{red}{\to 0}
\quad\mbox{as}\quad n \to \infty}^{\ds{\large\mbox{Laplace Method}}}
$$
