AN ANALYTICAL APPROACH TO A TRIGONOMETRIC INTEGRAL

Joseph Wiener The University of Texas – Pan American

The integral of sin^{2} x or cos^{2} x between 0 and \frac{π}{2} (or 0 and π) is usually calculated by changing the integrand according to the well–known half–angle formula. However, students with a phobia for trigonometry would prefer a simple geometric derivation of the answer suggested in [1]. The result is obtained immediately by noticing the equality of the areas of the regions under the graphs of sin^{2} x and cos^{2} x and by integrating the basic trigonometric identity sin^{2} x + cos^{2} x = 1. This idea is equivalent to the following analytical approach. Let

C = ∫0^{\frac{π}{2}} cos^{2} x dx, S = ∫0^{\frac{π}{2}} sin^{2} x dx.

Since the substitution x = \frac{π}{2} − u transforms either integral to the other, then C = S. Furthermore, integrating the basic identity between 0 and \frac{π}{2} yields C + S = \frac{π}{2} , which implies C = S = \frac{π}{4} .

Some students even try to use this method for integrals of sin^{4} x and cos^{4} x between 0 and \frac{π}{2} . From the start, things evolve as expected, the above substitution again shows the equality of the corresponding integrals, and a few students decide to integrate the “identity” sin^{4} x + cos^{4} x = 1. Doubts and questions quickly reveal the error, gradually lead to serious work, and raise new questions on further generalizations. Of course, Wallis’s integrals

C_{n} = ∫0^{\frac{π}{2}} cos^{2n} x dx, S_{n} = ∫0^{\frac{π}{2}} sin^{2n} x dx

are too hard for most students but may be suggested as a challenging exercise. The following method of evaluating C_{n} is new and unusual in that it reduces to a minimum the amount of trigonometric transformations and illustrates the interaction between integration and differentiation. It is easy to calculate the improper integral

∫_{0}^{∞} \frac{dy}{y^{2} + p} = \frac{π}{2} p^{− \frac{1}{2}} , p > 0.

Successively differentiating this relation with respect to p yields

∫_{0}^{∞} \frac{dy}{(y^{2} + p)^{n+1}} = \frac{1}{2} \frac{(2n)!}{4^{n}(n!)^{2}} πp^{− \frac{(2n+1)}{2}} .

Letting p = a^{2}, y = a tan x, we obtain the remarkable Wallis’s

formulas

C_{n} = S_{n} = \frac{(2n)!}{4^{n}(n!)^{2}} (\frac{π}{2} ).

\underline{Reference}

1. R. Euler, “A Geometric Approach to a Trigonometric Integral,” Missouri J. Math. Sci., Winter 1989, Vol. 1. No. 1, 28– 29.