I have been trying to evaluate this integral:

$$\int_0^\frac{\pi}{4} \frac{\tan^2(x)}{x^2+1} \,dx$$

According to the answer key that I have, the answer is supposed to be simply $1/3$. But, inputting it into Wolfram Alpha yields the decimal approximation of $\approx 0.156503$. I am trying to find an exact answer and all my attempts have failed.

I cannot see that any basic integration technique would work (i.e. u-substitution, integration by parts, etc), and I have tried to use symmetry to evaluate this (i.e. substitute $x=\frac{\pi}{4}-u$) but to no avail.

Thank you!

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    $\begingroup$ Are you sure is $\tan^2(x)$ and not $\arctan^2(x)$ or maybe $\tan^{-2}(x)$. I say it because $\frac1{1+x^2}$ happens to be the derivative of $\arctan(x)$. $\endgroup$ – Alejandro Nasif Salum Sep 10 '18 at 0:13
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    $\begingroup$ Probably a typo in the problem. Wolfram estimate looks correct. $\endgroup$ – Vasya Sep 10 '18 at 0:18
  • $\begingroup$ @AlejandroNasifSalum Yes, I double-checked, and this is the exact way the question is written (i.e. with no arctangent function). I believe the $1/3$ is definitely a typo, but I'm still unable to find a closed-form solution to this integral. $\endgroup$ – AJ123 Sep 10 '18 at 1:39
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    $\begingroup$ Possible duplicate of MIT 2015 Integration Question $\endgroup$ – James Arathoon Sep 11 '18 at 13:08

It's easy to see that $f(x)=\frac{\tan(x)^2}{x^2+1}$ is strictly increasing and convex on the integration interval $I=(0,\frac{\pi}{4})$. So on the interval $I$, the function $f$ assumes the maximum value in $x=\frac{\pi}{4}$: $f(\frac{\pi}{4})=\frac{\tan(\frac{\pi}{4})^2}{(\frac{\pi}{4})^2+1}=\frac{16}{16+\pi^2}$. Now construct a new function $g(x)=\frac{16}{16+\pi^2}x$. This is a line that pass trought the origin and the point $(\frac{\pi}{4},f(\frac{\pi}{4}))$. $f$ is convex on $I$ hence on the interval $I$: $$f(x)\leq g(x) \implies \int_{0}^{\frac{\pi}{4}}f(x)dx \leq \int_{0}^{\frac{\pi}{4}}g(x)dx = \int_{0}^{\frac{\pi}{4}}x\frac{16}{16+\pi^2}dx=\frac{\pi^2}{32+2\pi^2}\approx 0.19 < \frac{1}{3}$$ Finally $$\int_{0}^{\frac{\pi}{4}}f(x)dx < 0.19 < \frac{1}{3}$$


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