Problem 11961
(American Mathematical Monthly, Vol.124, February 2017) Proposed by M. Berindeanu (Romania).
Evaluate
Z π/2 0
sin(x) 1 +psin(2x)dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. By letting t = x − π/4, and√
2 sin(t) = sin(θ), we have that
I :=
Z π/2 0
sin(x)
1 +psin(2x)dx = Z π/4
−π/4
sin(t + π/4) 1 +psin(2t + π/2)dt
= 1
√2 Z π/4
−π/4
sin(t) + cos(t)
1 +pcos(2t) dt = 0 + 2
√2 Z π/4
0
cos(t) 1 +q
1 − 2 sin2(t) dt
= Z π/4
0
√2 cos(t)dt 1 +
q 1 − (√
2 sin(t))2
= Z π/2
0
cos(θ)dθ 1 + cos(θ)
= Z π/2
0
1 − 1
1 + cos(θ)
dθ = [θ − tan(θ/2)]π/20 =π 2 − 1.
