$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$ on 2015 MIT Integration Bee So one of the question on the MIT Integration Bee has baffled me all day today $$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$$ I have tried a variety of things to do this, starting with
Integration By Parts Part 1
$$\frac{\tan x-x}{1+x^2}\bigg\rvert_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\frac{-2x(\tan x -x)}{\left (1+x^2 \right )^2}\text{d}x$$ which that second integral is not promising, so then we try
Integration By Parts Part 2
$$\tan^{-1} x\tan^2 x\bigg\rvert_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}2\tan^{-1} x\tan x\sec^2 x\text{d}x$$ which also does not seem promising
Trig Substitution
$x=\tan\theta$ which results $$\int_{0}^{\tan^{-1}\frac{\pi}{4}}\tan^2 \left (\tan\theta\right )\text{d}\theta$$ which I think too simple to do anything with (which may or may not be a valid reason for stopping here) I had some ideas following this like power reducing $\tan^2 x=\frac{1-\cos 2x}{1+\cos 2x}$ which didn't spawn any new ideas. Then I thought maybe something could be done with differentiation under the integral but I could not figure out how to incorporate that. I also considered something with symmetry somehow which availed no results. I'm also fairly certain no indefinite integral exists. Now the answer MIT gave was $\frac{1}{3}$ but wolfram alpha gave $\approx$ .156503. Note
The integral I gave was a simplified version of the original here is the original in case someone can do something with it $$\int_{0}^{\frac{\pi}{4}}\frac{1-x^2+x^4-x^6...}{\cos^2 x+\cos^4 x+\cos^6 x...}\text{d}x$$ My simplification is verifiably correct, I'd prefer no complex analysis and this is from this Youtube Video close to the end.

In terms of the derivatives of the digamma function $\psi$ and the incomplete beta function $B$ we have
$$
\int_0^{\pi/4}\frac{\tan^2x}{1+x^2}dx
=\int_0^{\pi/4}\frac{1}{1+x^2}\left[
\sum_{k=1}^{\infty}
\frac{2 \psi ^{(2 k+1)}\left(\frac{1}{2}\right)}{\pi^{2k+2}(2 k)!}x^{2 k}
\right]dx\\
=\sum_{k=1}^{\infty}
\frac{2 \psi ^{(2 k+1)}\left(\frac{1}{2}\right)}{\pi^{2k+2}(2 k)!}
\int_0^{\pi/4}\frac{x^{2k}}{1+x^2}dx\\
=\sum_{k=1}^{\infty}
\frac{2 \psi ^{(2 k+1)}\left(\frac{1}{2}\right)}{\pi^{2k+2}(2 k)!}
\left(-\frac{i}{2} (-1)^k B_{-\pi^2/16}(k+\frac{1}{2},0)\right)\\
=-i\sum_{k=1}^{\infty}
\frac{(-1)^k\psi ^{(2 k+1)}\left(\frac{1}{2}\right)}{\pi^{2k+2}(2 k)!}
B_{-\pi^2/16}(k+\frac{1}{2},0).
$$
This sum converges rapidly: for large $k$, each term is about $1/4$ of the preceding term.
Summing the first few terms, we see that the result is definitely not $\frac13$: in Mathematica (see also here),
-I Sum[(-1)^k PolyGamma[2k+1, 1/2] Beta[-π^2/16, k+1/2, 0]/(π^(2k+2) (2k)!), {k, 30}] //N
(*    0.15650324569956275` - 3.530585719508859`*^-17 I    *)

