TITLE: MIT 2015 Integration Question
QUESTION [56 upvotes]: 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.

REPLY [4 votes]: To avoid the use of complex analysis as done above by @Przemo, could one simply use taylor series expansions and Cauchy products to obtain:
\begin{array}{rcl}
\displaystyle{\int_0^{\frac{\pi}{4}}\!\!\frac{\tan^2\!x}{1+x^2}dx} & = & \displaystyle{\int_0^{\frac{\pi}{4}}\!\!\tan^2\!x\left(1-x^2+x^4-x^6+\cdots\right)dx} \\[5mm]
& = & \displaystyle{\int_0^{\frac{\pi}{4}}\!\!\tan^2\!x\left(\sum_{k=0}^{\infty}(-1)^kx^{2k}\right)dx} \\[5mm]
& = & \displaystyle{\int_0^{\frac{\pi}{4}}\!\!\left(\sum_{k=0}^{\infty}\frac{B_{2k+2}(-4)^{k+1}(1-4^{k+1})}{(2k+2)!}x^{2k+1}\right)^{\!\!2}\left(\sum_{k=0}^{\infty}(-1)^kx^{2k}\right)dx} \\[5mm]
& = & \displaystyle{\int_0^{\frac{\pi}{4}}\!\!\left(\sum_{k=0}^{\infty}\sum_{l=0}^k\frac{B_{2l+2}B_{2k-2l+2}(-4)^{k+2}(1-4^{l+1})(1-4^{k-l+1})}{(2l+2)!(2k-2l+2)!}x^{2k+2}\right)\left(\sum_{k=0}^{\infty}(-1)^kx^{2k}\right)dx} \\[5mm]
& = & \displaystyle{\int_0^{\frac{\pi}{4}}\!\!\sum_{k=0}^{\infty}\sum_{l=0}^k\sum_{m=0}^l\frac{B_{2m+2}B_{2l-2m+2}(-4)^{l+2}(1-4^{m+1})(1-4^{l-m+1})}{(2m+2)!(2l-2m+2)!}(-1)^{k-l}x^{2k+2}dx} \\[5mm]
& = & \displaystyle{\sum_{k=0}^{\infty}\sum_{l=0}^k\sum_{m=0}^l\frac{B_{2m+2}B_{2l-2m+2}(-4)^{l+2}(1-4^{m+1})(1-4^{l-m+1})}{(2m+2)!(2l-2m+2)!}(-1)^{k-l}\int_0^{\frac{\pi}{4}}x^{2k+2}dx} \\[5mm]
& = & \displaystyle{\sum_{k=0}^{\infty}\sum_{l=0}^k\sum_{m=0}^l\frac{B_{2m+2}B_{2l-2m+2}(-4)^{l+2}(1-4^{m+1})(1-4^{l-m+1})}{(2m+2)!(2l-2m+2)!}(-1)^{k-l}\frac{\frac{\pi}{4}^{2k+3}}{2k+3}} \\[5mm]
\end{array}