The limit of $\int_{0}^{1}\frac{\sqrt{n}}{1+n\ln(1+x^2)}dx$ as $n\to\infty$ The task is to calculate $$\lim_{n\to\infty}\int_{0}^{1}\frac{\sqrt{n}}{1+n\ln(1+x^2)}dx$$ I tried various estimates I know to find the dominating integrable function and nothing worked. Does anyone have any idea? Is this even an application of DCT or something else? 
 A: DCT as Discrete Cosine Transform? I cannot understand what you are meaning.
Anyway, your integral converges towards $\frac{\pi}{2}$, since:
$$\int_{0}^{1}\frac{\sqrt{n}}{1+n\log(1+x^2)}\,dx \geq \int_{0}^{1}\frac{\sqrt{n}}{1+nx^2}\,dx = \arctan(\sqrt{n})=\frac{\pi}{2}+O\left(\frac{1}{\sqrt{n}}\right),$$
while the difference between the first and the second integral is bounded by:
$$\begin{eqnarray*}&&n^{3/2}(1-\log 2)\int_{0}^{1}\frac{x^4\,dx}{(1+nx^2)(1+n\log(1+x^2))}\\&\leq& n^{3/2}(1-\log 2)\int_{0}^{1}\frac{x^4 \,dx}{(1+n\log(1+x^2))^2}\\&\leq&n^{3/2}(1-\log 2)\int_{0}^{1}\frac{x^4 \,dx}{(1+nx^2-\frac{n}{2}x^4)^2}\end{eqnarray*}$$
where:
$$\frac{x^4}{(1+nx^2-\frac{n}{2}x^4)^2}\leq \frac{1}{n^2}\text{ for }x\in[0,1/\sqrt{n}],$$
$$\frac{x^4}{(1+nx^2-\frac{n}{2}x^4)^2}\leq \frac{1}{n^2(1-\frac{1}{2}x^2)^2}\leq\frac{4}{n^2}\text{ for }x\in[1/\sqrt{n},1],$$
hence:

$$\int_{0}^{1}\frac{\sqrt{n}}{1+n\log(1+x^2)}\,dx =\frac{\pi}{2}+O\left(\frac{1}{\sqrt{n}}\right).$$

A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Heuristically, when $\ds{n \ggg 1}$, the main contribution to the integral comes from $\ds{x \gtrsim 0}$ such that

\begin{align}
&\int_{0}^{1}{\root{n} \over 1 + n\ln\pars{1 + x^{2}}}\,\dd x
\sim\int_{0}^{1}{\root{n} \over 1 + nx^{2}}\,\dd x
=\arctan\pars{\root{n}} = {\pi \over 2} - \arctan\pars{1 \over \root{n}}
\\[3mm]&\stackrel{n \to \infty}{\to} \color{#66f}{\large{\pi \over 2}} \approx {\tt 1.5708}
\end{align}
