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 \(\left( {\frac{\pi }{4},\frac{1}{2}} \right)\).
![](https://i0.wp.com/www.raucci.net/wp-content/uploads/2021/09/f1-1024x868.png?resize=612%2C519&ssl=1)
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:
![](https://i0.wp.com/www.raucci.net/wp-content/uploads/2021/09/fx2-1024x768.png?resize=626%2C470&ssl=1)
![](https://i0.wp.com/www.raucci.net/wp-content/uploads/2021/09/social-1024x289.jpg?resize=1024%2C289&ssl=1)
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:
\(\int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt 2 }}x}}} = \frac{\pi }{4}\)
![](https://i0.wp.com/www.raucci.net/wp-content/uploads/2021/09/integral.png?resize=864%2C583&ssl=1)