$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$ 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.
 A: 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.
A: 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}
A: There must be some problem here.  Note that on $[0, \pi/4]$, $$\frac{\tan^2 x}{1+x^2} \le \frac{\tan^2 x}{1 + 0^2} = \tan^2 x,$$ therefore the definite integral is bounded above as follows:  $$0 \le \int_{x=0}^{\pi/4} \frac{\tan^2 x}{1+x^2} \, dx < \int_{x=0}^{\pi/4} \tan^2 x \, dx = 1 - \frac{\pi}{4} < \frac{1}{3}.$$  Mathematica is correct; the definite integral cannot be $1/3$.  I also watched the YouTube video (see time stamp 1:34:34), and you have transcribed the question correctly and performed the correct algebraic transformations.
Okay so at 1:39:04 they reveal the answer as $1/3$.  This is very, very obviously wrong.  The very first thought to enter my mind was to check the reasonableness of the result by choosing an appropriate bound.
Interestingly, one contestant at 1:38:42 answers with $\pi/16 \approx 0.19635$ which is remarkably accurate for a last-moment guess, certainly much closer to the mark than the official answer!
A: 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
$$I_n=\int_0^{\frac \pi 4} P_n\,dx$$
$$\left(
\begin{array}{cc}
n & I_n \\
 0 & 0.1614910 \\
 1 & 0.1509670 \\
 2 & 0.1565241 \\
 3 & 0.1565032 \\
\end{array}
\right)$$
A: 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    *)

A: Let $f(x)=\tan x-x=\sum_{n=2}^\infty a_nx^{2n-1}$ where $a_n=\frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}$ and $B_{2n}$ is the $2n$-th Bernoulli number. Then $f'(x)=\tan^2 x=\sum_{n=2}^\infty (2n-1)a_nx^{2n-2}$. Hence,
$$\int_0^{\pi/4}\frac{\tan^2x}{x^2+1}dx=\sum_{n=2}^\infty (2n-1)a_nb_n$$
where
$$b_n=\int_0^{\pi/4}\frac{x^{2n-2}}{x^2+1}dx.$$
A: I read that in the final of this competition only four minutes are given for one integral.
It is madness to try to calculate the given integral in that time.
Since the competition takes place at MIT then i suppose that simple approximations to get quick estimations of integrals is approved.
In given case let's use the very crude  but the simplest approximations to approximate the integrand.
$$\tan x \approx \frac{4x}{\pi}$$
$$\frac{1}{1+x^2}\approx 1-\frac{4\pi x}{16+\pi^2}$$
$$0\leqslant x \leqslant  \frac{\pi}{4} $$
They are chosen in such a way that the errors they cause are directed oppositely and partially these errors cancel one another out.
i.e.
$$\tan x < \frac{4x}{\pi}$$
$$\frac{1}{1+x^2} > 1-\frac{4\pi x}{16+\pi^2}$$
Then simple integration gives
$$I\approx \frac{\pi}{48} \frac{64+\pi^2}{16+\pi^2}=0.1868...$$
The exact value of the original integral is $I=0.1565...$ so that the deviation from the exact value is about $0.03$
So we avoided the complicated computations and still gain insight into the  value of the given integral.
There was a desire to use complex analysis to calculate this integral.
I think that if it is possible, then at the cost of very complicated calculations.
