A clever integral

The integrand doesn’t have an obvious antiderivative and it’s not clear how one would go about computing it. So what can we do? In this case, the limits of integration play a key role. Let’s call our problematic integral $J_1$. Note that:

\[
J_1 = \int_0^{\pi/2} \frac{\mathrm{d}x}{1 + (\tan x)^{\sqrt{2}}} = \int_0^{\pi/2} \frac{(\cos x)^{\sqrt{2}}}{(\cos x)^{\sqrt{2}} + (\sin x)^{\sqrt{2}}} \,\mathrm{d}x
\]

No progress yet, but the new form of $J_1$ suggests a certain symmetry. Specifically, define:

\[
J_2 = \int_0^{\pi/2} \frac{(\sin x)^{\sqrt{2}}}{(\cos x)^{\sqrt{2}} + (\sin x)^{\sqrt{2}}} \,\mathrm{d}x
\]

Two key observations. First, $J_1 + J_2 = \pi/2$. This is clear because the integrands of $J_1$ and $J_2$ sum to 1. Second, $J_1=J_2$. This is clear because if we make the substitution $x \mapsto \frac{\pi}{2} – x$, then $\cos$ and $\sin$ trade places and the integral remains unchanged. These facts together imply that

\[
J_1 = J_2 = \frac{\pi}{4}
\]

Here is a plot showing the integral represented as an area under the curve. This area is precisely half the area of the entire rectangle.

Note also that the $\sqrt{2}$ was a red herring. We could have replaced it by some other real number and the value of the integral would be unchanged!

We managed to evaluate the integral without ever computing the antiderivative. This isn’t uncommon — for example, a standard integral that shows up when working with normal distributions is:

\[
\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}x = \sqrt{\pi}
\]

This integral can be evaluated even though $e^{-x^2}$ doesn’t have an antiderivative… but that’s a topic for another post!