Convergence to $\ln(n)$ Is there any hint for $\,\displaystyle\lim_{n \to \infty} \int_{0}^{\pi/2}\!\!\!\frac{n\cos x}{\sqrt{1+n^2x^2}}\,\mathrm dx\;$?
I've tried with $n \approx 9000000000000000000$ and find that it seems to converge to $\ln(n)$ but no clue to prove that.
E.g. $n=9000000000000000000$, the integral is $44.2316884183270$ (while $\ln(n) = 43.6437562512290$). $n=90000000000000000$, the integral is $39.6265182323389$ (while $\ln(n) = 39.0385860652409$)
 A: Let's denote $\displaystyle I(n)=n\int_0^\frac{\pi}{2}\frac{\cos x}{\sqrt{1+n^2x^2}}dx$.
Then, integrating by part,
$$=\int_0^\frac{\pi}{2}\frac{\cos x}{\sqrt{\frac{1}{n^2}+x^2}}dx=\cos x\ln\left(x+\sqrt{\frac{1}{n^2}+x^2}\,\right)\,\bigg|_0^\frac\pi2+\int_0^\frac{\pi}{2}\ln\left(x+\sqrt{\frac{1}{n^2}+x^2}\,\right)\sin x\,dx$$
The second term converges at $n\to\infty$. Therefore, with the accuracy up to $o(1)$,
$$I(n)=\ln n+\int_0^\frac{\pi}{2}\ln(2x)\sin x\,dx+o(1)=\ln n+\ln 2+\int_0^\frac{\pi}{2}\ln x \,\sin x\,dx+o(1)$$
$$=\ln n+\ln 2-\gamma+\operatorname{Ci}\big(\frac\pi2\big)+o(1)$$
Numeric check with WolframAlpha
$\displaystyle n=100 \quad I(100)=\color{blue}{5.193}\,241...\qquad \ln (100)+\ln 2+\int_0^\frac{\pi}{2}\ln x \,\sin x\,dx=\color{blue}{5.193}\,102... $

With some additional efforts next asymptotic terms can be found:
$$I(n)=n\int_0^\frac{\pi}{2}\frac{\cos x}{\sqrt{1+n^2x^2}}dx=\ln n+\ln 2+\int_0^\frac{\pi}{2}\ln x\,\sin x\,dx$$
$$+\frac{1}{4n^2}\Big(\ln n+\ln2+\int_0^\frac{\pi}{2}\ln x\,\sin x\,dx+1-\frac{2}{\pi}\Big)+o\big(\frac{1}{n^2}\big)$$
Numeric check with WolframAlpha
$\displaystyle n=100 \quad I(100)=\color{blue}{5.193\,241\,26}6\,17...\quad \text{approximation}=\color{blue}{5.193\,241\,26}5\,15... $
A: $$I_n:=\int_0^{\pi/2}\frac{n\cos x}{\sqrt{1+n^2x^2}}\,\mathrm dx=J_n-K_n$$ where $$J_n:=\int_0^{\pi/2}\frac{\mathrm dx}{\sqrt{x^2+\frac1{n^2}}}=\sinh^{-1}\frac{n\pi}2=\ln n+\ln\left(\frac\pi2+\sqrt{\frac1{n^2}+\frac{\pi^2}4}\right)$$
and (by dominated convergence)
$$K_n:=\int_0^{\pi/2}\frac{1-\cos x}{\sqrt{x^2+\frac1{n^2}}}\,\mathrm dx\longrightarrow K:=\int_0^{\pi/2}\frac{1-\cos x}x\,\mathrm dx$$
($=-{\rm Ci}\frac\pi2+\gamma+\ln\frac\pi2\approx0.5568$)
hence
$$I_n-\ln n\longrightarrow-K+\ln\pi$$
($={\rm Ci}\frac\pi2-\gamma+\ln2$ $\approx0.588$).
