<s>[INST] $\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$ 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.
 [/INST] This is not the whole answer to this question but it is something new that wasn't mentioned here so I thought I would post it. Let us denote the unknown integral by $J$. Then substituting for $y= \tan(x)$ we get:
\begin{equation}
J = \int\limits_0^1 \frac{y^2}{1+\arctan(y)^2} \cdot \frac{1}{1+y^2} d y = 
\int\limits_0^1 \frac{1}{1+\arctan(y)^2} d y - \int\limits_0^1 \frac{1}{1+\arctan(y)^2} \cdot \frac{dy}{1+y^2}=
\left(\sum\limits_{n=0}^\infty (-1)^n \int\limits_0^1 \arctan(y)^{2 n} d y\right) - \arctan(\frac{\pi}{4})
\end{equation}
Now, clearly the series converges which follows from the estimate $\arctan(y) < y$ for $y\in(0,1)$. Integrating by parts we have:
\begin{eqnarray}
\int\limits_0^1 \arctan(y)^2 d y &=& (\frac{\pi}{4})^2+ \frac{\pi}{4} \log(2) - G \\
\int\limits_0^1 \arctan(y)^{2 n} d y &=& \cdots
\end{eqnarray}
where $G$ is the Catalan constant. Now, of course the real challenge is to calculate the last integral above in closed form. This requires more work. However I have a feeling that this integral can be reduced to poly-logarithms for arbitrary values of $n$ and as such does have a "closed form". 
We use the following identities:
\begin{eqnarray}
 [\arctan(y)]^2 &=& \frac{1}{2} \sum\limits_{n=0}^\infty (-1)^n \frac{(y^2)^{n+1}}{n+1} \cdot \left( \Psi(-1/2) + \Psi(n+3/2)\right) \\
&=& -\frac{1}{4} \log ^2\left(\frac{y+i}{y-i}\right)+\frac{1}{2} i \pi  \log \left(\frac{y+i}{y-i}\right)+\frac{\pi ^2}{4}
\end{eqnarray}
Now, raising the identity above to the $n$th power and integrating we readily see that the only non-trivial integral we are dealing is, is the following:
\begin{eqnarray}
{\mathcal A}_n &:=& \int\limits_0^1 [\log(\frac{x+\imath}{x-\imath})]^n dx \\
&=& (-2 \imath) \int\limits_{-1}^{\imath} \frac{ [\log(u)]^n}{(1-u)^2} d u \\
&=&(-\imath)(\imath \pi)^n + (1+\imath) (\imath \frac{\pi}{2})^n + \\
&&(-\imath)n! (-1)^n \left((\log(2)+\imath \frac{\pi}{2}) 1_{n=1} + 2(S_{1,n-1}(2) - S_{1,n-1}(1-\imath)) 1_{n > 1}\right)
\end{eqnarray}
where the path in middle integral is a quarter of a unit circle starting at $-1$ and ending at $\imath$. The quantities $S_{1,n-1}()$ in the bottom formula are the Nielsen generalized poly-logarithms. Here $n\ge 1$. Combining the two identities above we obtain the following identity :
\begin{eqnarray}
&&\int\limits_0^1 \arctan(y)^{2 n} dy = \\
&&
 \left(\frac{\pi^2}{16}\right)^n + \\
&&  Re\sum\limits_{0\le p_1 \le p_2 \le n} 
\frac{n! (2(p_2-p_1)+n-p_2)!}{p_1! (p_2-p_1)!(n-p_2)!} 
\left(\frac{\pi^2}{4}\right)^{p_1} \left(\frac{1}{4}\right)^{p_2-p_1} \left(\frac{\pi}{2}\right)^{n-p_2} \cdot \\
&&
\left(
0 \cdot 1_{n+p_2-2 p_1=0} + 
(\log(2)) \cdot 1_{n+p_2-2 p_1=1} + 
2 (-\imath)^{n+p_2-2 p_1+1} 
(S_{1,n+p_2-2 p_1-1}(2) - S_{1,n+p_2-2 p_1-1}(1-\imath)) \cdot 
1_{n+p_2-2 p_1>1}  
\right)
\end{eqnarray}
valid for $n=1,2,\cdots$. This concludes the calculation.
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