A simple integral

The problem is to compute the following integral


There is a way to find this integral actually without computing integrals at all. Well, almost.


  1. R. Honsberger, More Mathematical Morsels, MAA, New Math Library, 1991, p. 136-137
  2. G.-C. Rota, Indiscrete Thoughts, Birkhauser, 1997

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Copyright © 1996-2017 Alexander Bogomolny

Compute the followng integral



First of all, observe that


Therefore, summing up the two and using the well known identity, we get that their sum equals 2π. In particular,

\(\displaystyle\int_{0}^{2\pi}\mbox{sin}^{2}(x)dx=\pi .\)

This solves the problem. However, several integrals may be obtained with no additional effort. For one, the second integral above is also π. Now, since the period of sin²(x) is actually π, we easily get another integral

\(\displaystyle\int_{0}^{\pi}\mbox{sin}^{2}(x)dx=\pi /2.\)

Finally, taking into account the symmetry of \(\mbox{sin}\), \(\mbox{sin}(\pi - x) = \mbox{sin}(x)\), we also obtain

\(\displaystyle\int_{0}^{\pi /2}\mbox{sin}^{2}(x)dx=\pi /4.\)

On a flippant note, every time we halved the interval of integration, the integral got halved as well. Would it be right to expect that the next time we'll get an integral equal to one eighth of \(\pi \)?

In a similar vein, consider the following integral:

\(\displaystyle I=\int_{0}^{\pi /2}\frac{\mbox{sin}^{25}(x)}{\mbox{sin}^{25}(x)+\mbox{cos}^{25}(x)}dx\)

The values taken by the integrand \(f(x)\), as \(x\) runs from \(0\) to \(\pi /2\), are exactly the same as taken on by \(f(\pi /2-x)\) on the same interval, except that they are generated in reverse order. Consequently, the integrals of the two functions are equal:

\(\displaystyle I=J=\int_{0}^{\pi /2}\frac{\mbox{sin}^{25}(\pi /2-x)}{\mbox{sin}^{25}(\pi /2-x)+\mbox{cos}^{25}(\pi /2-x)}dx=\int_{0}^{\pi /2}\frac{\mbox{cos}^{25}(x)}{\mbox{cos}^{25}(x)+\mbox{sin}^{25}(x)}dx\)

Summing up,

\(\displaystyle I+J=\int_{0}^{\pi /2}1dx.\)

Since \(I = J\), so \(I = \pi /4\). What does exponent \(25\) have to do with that? Nothing! For, the integrand falls into an even more general framework, by far.

Some sums, too, can also be found without employing any calculations.

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Copyright © 1996-2017 Alexander Bogomolny


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