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Taking advantage of symmetry…


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

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:

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

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