$\displaystyle \int_0^{\pi} \cos(\tan(x)) ~dx ~= \frac{\pi}{e} $ ???
Huh?
$\displaystyle cos(tan(x)) dx$ = $\displaystyle cos(u) du$
$\displaystyle du = sec^2x dx $ which doesn't make sense.
How does $\displaystyle cos(tan(x)) = cos(x)/(1+x^2)$?
Please clarify. Maybe there's some trig information I'm missing or something.