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College Math Journal Problem 667 Solution.
Peter Y. Woo, 2/10/2000Biola University, La Mirada, Ca.
Problem. Evaluate
ò p0 [2 + 2 cos x - cos (n-1)x - 2 cos n x - cos (n+1)x ] / (1- cos 2 x) dx .
Proof. Let I = the integrand. Let sm and cm denote sin m x and cos m x . The numerator of I
= 4 c(½)2 - 2 c(½) (c(n-(½)) + c(n + (½)))
= 4 c(½)2 (1 - cn ) = 8 c(½)2 sn/22.
The denominator = 2 s12 = 8 s(½)2 c(½)2 .
Hence I = [sn/2 / s(½)]2 .
Case 1: Suppose n is odd, n = 2 m + 1. From the identity
2 (c1 + c2 + . . . + cm) s(½) = - s(½) + sm+(½) ,
we get I = [1 + 2 ( c1 + c2 + . . . + cm)]2
= 1 + 4(c12 + c22 + . . . + cm2) + å0 £ h < k £ m ahk ch ck .
Now ò p0 ch ck dx = 0 whenever h ¹ k. Hence ò p0 I dx
= ò p0 1 + 2[ (1 + c2) + (1 + c4) + . . . + (1 + c2m)] dx
= ò p0 1 + 2 m dx = n p .
Case 2: Suppose n is even, n = 2 m. From the identity
2 s(½) (c(½) + c1+(½) + . . . + cm-(½)) = sm , we get
ò p0 I dx = 4 ò p0 (c(½) + c1+(½) + . . . + cm-(½))2 dx
= 4 ò p0 c(½)2 + c1+(½)2 + . . . + cm-(½)2 dx
[because ò p0 ch+(½) ck+(½) dx for any integers h ¹ k still = 0 . ] Hence
ò p0 I dx = ò p0 2 [(1+c1) + (1+c3) + . . . + (1+c2m-1)] dx
= 2 m p = n p .
