Proving $\lim\limits_{n\to\infty}\int_0^{\pi/4} \tan^n{x}\,dx=0$ How would you prove that $\displaystyle\lim\limits_{n\to\infty}\int_0^{\pi/4} \tan^n{x}\,dx=0$.
It is obvious if you see the graph of $\tan^n{x}$ on $(0, \pi/4)$ as $n$ increases but i'm looking for a more algebraic way. 
This result is for connecting the power reduction formula for $\displaystyle\int_0^{\pi/4} \tan^n{x}\,dx$ to Leibniz formula for $\pi$.
 A: Squeezing is straightforward:
$$ 0\leq \int_{0}^{\pi/4}\tan^n(x)\,dx \stackrel{x\mapsto\arctan u}{=}\int_{0}^{1}\frac{u^n}{1+u^2}\,du \leq \int_{0}^{1}u^n\,du = \frac{1}{n+1}.$$
A: Let $0<\varepsilon<\frac{\pi}{4}$. Write the integral
$$I_n=\int_0^{\pi/4}\tan^n x\,\mathrm dx=\int_0^{\pi/4-\varepsilon}\tan^n x\mathrm dx+\int_{\pi/4-\varepsilon}^{\pi/4}\tan^n x\,\mathrm dx=J_n+K_n$$
First notice the limit of $I_n$ exists, because it's decreasing and positive, since $0\le\tan^{n+1}x\le\tan^{n}x$.
Since the function $x\to\tan^n x$ is increasing, the first term $J_n$ is bounded by:
$$J_n\le(\frac{\pi}{4}-\varepsilon)\tan^n(\frac{\pi}{4}-\varepsilon)\le\frac{\pi}{4}\beta^n$$
With $0<\beta<1$.
Since $\tan^n x\le1$, The second term $K_n$ is bounded by:
$$K_n\leq\varepsilon$$
Hence, for all $\varepsilon$, there is a $\beta$ such that $0<\beta<1$ such that for all $n$,
$$0\le I_n\le \frac{\pi}{4}\beta^n+\varepsilon$$
Hence, for all $\varepsilon$,
$$0\le\lim_{n\to\infty} I_n\le\varepsilon$$
So
$$\lim_{n\to\infty} I_n=0$$
A: More than likely, this is too complex.
The antideirvative can be computed using hypergoemetric functions
$$I_n=\int \tan ^{n}(x)\,dx=\frac{\tan ^{n+1}(x)}{n+1} \, _2F_1\left(1,\frac{n+1}{2};\frac{n+3}{2};-\tan   ^2(x)\right)$$ making
$$J_n=\int_0^{\frac \pi 4} \tan ^{n}(x)\,dx=\frac{1}{4} \left(\psi \left(\frac{n+3}{4}\right)-\psi
   \left(\frac{n+1}{4}\right)\right)$$ Using the asymptotics
$$\psi(z)=\log \left({z}\right)-\frac{1}{2 z}-\frac{1}{12 z^2}+\frac{1}{120
   z^4}+O\left(\frac{1}{z^6}\right)$$ and continuing using Taylor series
$$J_n=\frac{1}{2 n}-\frac{1}{2 n^3}+O\left(\frac{1}{n^5}\right)$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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With
  Laplace's Method:

\begin{align}
\int_{0}^{\pi/4}\tan^{n}\pars{x}\,\dd x & =
\int_{0}^{\pi/4}\tan^{n}\pars{{\pi \over 4} - x}\,\dd x =
\int_{0}^{\pi/4}\exp\pars{n\ln\pars{\tan\pars{{\pi \over 4} - x}}}\,\dd x
\\[5mm] &
\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\int_{0}^{\infty}\expo{-2nx}\bracks{1 -{4n \over 3}\,x^{3} + \mrm{O}\pars{nx^{5}}}\dd x
\\[5mm] & =
{1 \over 2n}  - {1 \over 2n^{3}} + \mrm{O}\pars{1 \over n^{5}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,\bbx{\large 0}
\end{align}
