$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$ 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.

Six years too late but just for the fun of approximations
Knowing the series expansion of the integrate, it is simple to build $P_n$, the $[2n+2,2n]$ corresponding Padé approximant.
$$P_n=x^2\,\frac{1+\sum_{k=0}^n a_k\,x^{2k}}{ 1+\sum_{k=0}^n b_k\,x^{2k}}$$ which can in turn write
$$P_n=\frac{a_n}{b_n} x^2\frac{\prod_{k=0}^n (x^2-r_k) } {\prod_{k=0}^n (x^2-s_k) }$$ Now, partial fraction decomposition and integration for closed form approximations. For example
$$P_1=x^2\,\frac{1+\frac 95 x^2 } {1+\frac {32}{15} x^2 }$$
$$I_1=\frac{3}{8192}\left(50 \pi +12 \pi ^3-25 \sqrt{30} \tan ^{-1}\left(\pi\sqrt{\frac{2}{15}}    \right)\right)$$
Some decimal representation of results for
$$I_n=\int_0^{\frac \pi 4} P_n\,dx$$
$$\left(
\begin{array}{cc}
n & I_n \\
 0 & 0.1614910 \\
 1 & 0.1509670 \\
 2 & 0.1565241 \\
 3 & 0.1565032 \\
\end{array}
\right)$$
