For $n\geq 2$, define the sequence $\{x_{n}\}$ by $$x_{n}=\frac{1}{2\pi}\int_{0}^{\pi/2}\tan^{\frac{1}{n}}t \ dt$$ Then prove that the sequence $\{x_{n}\}$ converges to $1/4$.
My Attampt: I think it is a application of Dominated convergence theorem, But I am not sure, give some Hints.
On
Under the interchange $x\leftrightarrow\frac{\pi}{2}-x$ consider the integral
$$I = \frac{1}{2\pi}\int_0^{\frac{\pi}{2}}\cot^{\frac{1}{n}}t\:dt$$
then we have on $\left(0,\frac{\pi}{2}\right)$
$$t < \tan t \implies t^{-\frac{1}{2}}> t^{-\frac{1}{n}} > \cot^{\frac{1}{n}}t$$
for $n\geq 2$. This gives
$$I < \frac{1}{2\pi}\int_0^{\frac{\pi}{2}}\frac{dt}{\sqrt{t}} = \frac{1}{\sqrt{2\pi}}$$
thus you can take the limit inside the integral by dominated convergence, with the dominating function
$$g(t) = \frac{1}{\sqrt{\frac{\pi}{2}-t}}$$
Divide the integral into two parts, from $0 < t < {\pi \over 4}$ and ${\pi \over 4} \leq t < {\pi \over 2}$. On the first interval one has $\tan t < 1$ and on the second one has $\tan t \geq 1$. Hence the functions $\tan^{1 \over n} t$ are increasing in $n$ on the left interval and decreasing in $n$ on the right interval. This should make it easier to find the right convergence theorem(s) to use on a given interval.