Ways to evaluate $\lim_{n\to\infty}\int_0^{\pi/2}\log \left( \frac{1}{n} +\tan^2x\right) dx$, as zero When I've computed with Wolfram Alpha the definite integral $$\int_0^{\pi/2}\log \left( \frac{1}{n} +\tan^2x\right) dx,$$ (see the code integrate log(1/n+tan^2(x))dx, from x=0 to x=pi/2) then I said myself that should be very easy to get a result from this, since the $\lim_{n\to\infty}$ of our RHS is zero. It is easy to see if we write it as $$\lim_{n\to\infty}\frac{1}{2}\pi \left( \log\left(\frac{n-1}{n}\right)+2\coth^  {-1}(\sqrt{n})\right) ,$$ and then we  calculate the limit of the second summand by using the definiton of the hyperbolic cotangent.
And now

Question. Can you to prove by means different calculations and justifications that $$\lim_{n\to\infty}\int_0^{\pi/2}\log \left( \frac{1}{n} +\tan^2x\right) dx=0?$$ Are required those calculations or hints in your claims and how you justify your claims. Thanks.

 A: We need that $\log (\sin x)$ is Lebesgue integrable over $\bigl[0,\frac{\pi}{2}\bigr]$. Since $\frac{2}{\pi}x \leqslant \sin x \leqslant 1$ on that interval, the integrability follows from that of $\log x$. Since $\cos x = \sin \bigl(\frac{\pi}{2} - x\bigr)$, the integrability of $\log (\cos x)$ is then clear. Next we note
$$\tan^2 x < \frac{1}{n} + \tan^2 x \leqslant 1 + \tan^2 x = \frac{1}{\cos^2 x},$$
and by the monotonicity of the logarithm
$$2\bigl(\log(\sin x) - \log (\cos x)\bigr) = \log (\tan^2 x) \leqslant \log \bigl(\tfrac{1}{n} + \tan^2 x\bigr) \leqslant -2\log (\cos x).$$
We thus have a pointwise convergent sequence of measurable functions sandwiched between two integrable functions, hence all these functions are integrable, and by the dominated convergence theorem
$$\lim_{n \to\infty} \int_0^{\pi/2} \log \bigl(\tfrac{1}{n} + \tan^2 x)\,dx = \int_0^{\pi/2} \log (\tan^2 x)\,dx = 2\int_0^{\pi/2} \log (\sin x) - \log (\cos x)\,dx = 0$$
since
$$\int_0^{\pi/2} \log (\cos x)\,dx = \int_0^{\pi/2} \log \bigl(\cos \bigl(\tfrac{\pi}{2} - u\bigr)\bigr)\,du = \int_0^{\pi/2} \log (\sin u)\,du.$$
