Asymptotic expansion of $ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $ I'm trying to compute the asymptotic expansion of
$$ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $$
Here is what I've done:
Change of variable $$ t= \tan x $$
$$ I_n = \int_0^1 \frac{t^n \mathrm dt}{1+t^2} = \int_0^1 \frac{(1-t)^n \mathrm dt}{t^2-2t+2} $$
Change of variable
$$ t=\frac{x}{n}$$
$$ I_n = \frac{1}{2n}\int_0^n \frac{\left(1-\frac{x}{n}\right)^n \mathrm dx}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}$$
Taylor expansions:
$$ \left(1-\frac{x}{n}\right)^n = e^{-x} \left(1-\frac{x^2}{2n}+\frac{3x^4-8x^3}{24n^2}+\mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
$$ \frac{1}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)} = 1+\frac{x}{n}+\frac{x^2}{2n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) $$
$$ \frac{\left(1-\frac{x}{n}\right)^n }{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}=e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
So 
$$ I_n = \frac{1}{2n} \int_0^n e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right) \mathrm dx $$
$$ I_n = \frac{1}{2n} \left(1+\frac{1}{n}+\frac{1}{2n^2}\times 2-\frac{1}{2n}\times 2 - \frac{1}{2n^2} \times 6 + \frac{1}{8n^2} \times 24 - \frac{1}{3n^2} \times 6+ \mathcal{O} \left(\frac{1}{n^3} \right) \right) $$
$$ I_n = \frac{1}{2n}-\frac{1}{2n^3}+\mathcal{O} \left(\frac{1}{n^4} \right)$$
For example Wolfram gives:
$$ 1-1000^2+ 2\times1000^3\int_0^{\pi/4} \tan(x)^{1000} \mathrm dx \approx 4.9\times 10^{-6}$$
I'm quite sure of my work, I would just like to know if everything is correct!

I couldn't follow what you were doing, but I do see that you have the right answer to first order.  I can, however, point to a simpler way based on Laplace's Method.
First of all, write the integral as an exponential whose maximum is at $x=0$:
$$I_n = \int_0^{\pi/4} dx \: e^{n \log{\tan{\left ( \frac{\pi}{4} - x \right )}}} $$
As $n \rightarrow \infty$, the integral is dominated by its contributions from where the integrand is a maximum, or at $x=0$.  We may then approximate $\log{\tan{\left ( \frac{\pi}{4} - x \right )}}$ by its lower-order terms in its Taylor expansion about $x=0$.
To first order, note that $\log{\tan{\left ( \frac{\pi}{4} - x \right )}} \sim -2 x$ as $x \rightarrow \infty$.  We may then write
$$I_n \sim \int_0^{\pi/4} dx \: e^{-2 n x} \: \: (n \rightarrow \infty)$$
We may also may as well send that upper integration limit to $\infty$, as, again, the contributions to the integral away from $x=0$ are exponentially small.  The leading order term in the expansion is then
$$I_n \sim \frac{1}{2 n} \: \: (n \rightarrow \infty)$$
You may also derive higher order terms from the higher-order terms in the Taylor expansion.  The integrals that result will come from Taylor expanding the exponential piece containing these higher-order terms, which will result in Gamma-function-like integrals which will bring in decreasing powers of $n$ in the asymptotic expansion.
EDIT
I will do out the next order.  To second order:
$$\log{\tan{\left ( \frac{\pi}{4} - x \right )}} =-2 x -\frac{4}{3} x^3 +O(x^4) $$
The corrected expansion is then
$$I_n \sim \int_0^{\infty} dx \: e^{-2 n x} \left ( 1 - \frac{4}{3} n x^3 \right ) \: \: (n \rightarrow \infty)$$
Doing this next integral as well, we find that
$$I_n \sim \frac{1}{2 n} - \frac{1}{2 n^3} \: \: (n \rightarrow \infty)$$
in agreement with your result.
