Problem 11457
(American Mathematical Monthly, Vol.116, October 2009) Proposed by M. L. Glasser (USA).
For real numbersa and b with0 ≤ a ≤ b, find
Z b
x=a
arccos x
p(a + b)x − ab
! dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We will show that
Z b
x=a
arccos x
p(a + b)x − ab
!
dx=π(a − b)2 4(a + b) .
We assume that b > a ≥ 0 (otherwise it trivially holds). Let α = a + b > 0, β = −ab < 0 and consider the function
F(x) = αx+ β α arccos
x
√αx+ β
+α2+ 4β
4α arcsin 2x − α pα2+ 4β
!
−1 2
pαx+ β − x2.
Note thatpα2+ 4β = |b − a| = b − a 6= 0. Moreover F is a differentiable function defined in (a, b) and it is to verify that the F′(x) = f (x) for x ∈ (a, b). Hence the integral is given by
F(b) − F (a) = π(a − b)2 4(a + b) .
