<s>[INST] Existence of $\lim_{k\to +\infty}\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\,\mathrm{d}x$ 
Prove that existence of limit $$\lim_{k\to +\infty}\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\,\mathrm{d}x$$ in $\mathbb{R}$.

Attempt. 
Let $A_k$ be the $k$-th term of the sequence. Then $(A_k)$ is increasing. If we proved that it is upper bounded, then we would have convergence. But the estimation:
$$\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\,\mathrm{d}x\leqslant \frac12\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-\sin^2x}}\,\mathrm{d}x=+\infty,$$
although corrent, seems to overestimate the integral.
Thanks for the help.
 [/INST] How about this?  
For fixed $x \in (0,\pi/2)$, the integrand
$$
\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}
$$
increases as $k$ increases. The integrand is nonnegative.  So by the monotone convergence theorem,
$$
\lim_{k\to\infty}\int_0^{\pi/2}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\;dx \\=
\int_0^{\pi/2}\lim_{k\to\infty}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\;dx \\=
\int_0^{\pi/2}\;\frac{\sin^2x}{2\sqrt{1-\sin^2x}}\;dx = +\infty
$$
</s>