Taking advantage of symmetry…

Estimate:

$\int\limits_0^{\frac{\pi }{2}} {\frac{{{d} x}}{{1 + {{\tan }^{\sqrt 2 }}x}}}.$

Solution: The problem cannot be evaluated by the usual techniques of integration; that is to say, the integrand does not have an antiderivative.

The problem can be handled if we happen to notice thet the integrand is symmetric about the point $\left( {\frac{\pi }{4},\frac{1}{2}} \right)$.

To show this is so, let

$f(x) = \frac{1}{{1 + {{\tan }^{\sqrt 2 }}x}}$

It suffices to show that

$f(x)+f(\pi /2 -x) = 1 \qquad \forall x \in \left[0, \pi/2\right]$

Note: The diagram of $f(\pi / 2 – x)$ is shown below:

It follow, from the symmetry just proved, that the area under the curve in $\left[0, \pi/2\right]$ is one-half area of rectangle, i.e. $\frac{1}{2} \frac{\pi}{2}= \frac{\pi}{4}$. So:

$\int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt 2 }}x}}} = \frac{\pi }{4}$