Asymptotics of the integral $\int_0^\pi x^n \sin(x)dx $ I am going through de bruijn's book on asymptotic methods. In the end of a chapter on Laplace's method for integrals, there is an exercise to show the following asymptotic:
$$\int_0^\pi x^n\sin(x)dx\sim \frac{\pi^{n+2}}{n^2}, n\to\infty $$
I couldn't relate this to the examples in the chapter, where he dealt mainly with integrals of the form $\int_I e^{-tx^2}f(x)dx $, where $t>0$ a real number, and where that width of the interval contribuiting the most for the result was small (here it is of constant length to my understanding, $[1,\pi]$). However I manged to show (using the obviouse bound $\sin(x)<x$) a weaker result. I tried to read through the chapter again, but I have no idea how to do better here.
I would very appreciate any hints or sketches of solution.
 A: For any $x\in(0,\pi)$ we have
$$ \frac{\sin x}{x(\pi -x)} = \frac{1}{\pi}+K x(\pi-x),\qquad K\in\left[\frac{1}{\pi^3},\frac{4(4-\pi)}{\pi^4}\right] \tag{1}$$
hence
$$ \int_{0}^{\pi}x^n\sin(x)\,dx = \frac{\pi^{n+2}}{(n+2)(n+3)}+\pi^n O\left(\frac{1}{n^3}\right)\tag{2}$$
due to $\int_{0}^{\pi}x^\alpha(\pi-x)^{\beta}\,dx = \pi^{\alpha+\beta+1}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\pi}x^{n}\sin\pars{x}\,\dd x} =
\int_{0}^{\pi}\pars{\pi - x}^{n}\sin\pars{x}\,\dd x
\\[5mm] = &\
\pi^{n + 1}\int_{0}^{1}\pars{1 - x}^{n}\sin\pars{\pi x}\,\dd x
\\[5mm] = &\
\pi^{n + 1}\int_{0}^{1}\exp\pars{n\ln\pars{1 - x}}
\sin\pars{\pi x}\,\dd x
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&
\pi^{n + 1}\int_{0}^{\infty}\expo{-nx}\pars{\pi x}\dd x =
\bbx{\pi^{n +2} \over n^{2}}
\end{align}
