I was recently reminded of this problem from one of my favorite books: Problem-Solving Through Problems. The problem originally appeared in the 1980 Putnam Competition.

Evaluate the following definite integral.

\[

\int_0^{\pi/2} \frac{\mathrm{d}x}{1 + (\tan x)^{\sqrt{2}}}

\]

The solution:

[Show Solution]

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.

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!

Hey Laurent!

This is a very nice problem; thanks! In a similar vein, on Math.SE I found this IMO training problem that you may like:

http://math.stackexchange.com/questions/741580/integral-int-pi-2-pi-2-frac12007x1-cdot-frac-sin2008x-si?rq=1

John