tanx = t,
∴sinx = 2t/(1 t^2)
dx = dt/(1 t^2)
dx = dt/(1 t^2)
p>
∴∫(0, π/2)(senx)^2/[1 (senx)^2)
=∫(0, ∞)t^2/[2( t ^2) 1][(t^2) 1]dt
Nota:
t^2 =[2(t^2) 1]-[(t^ 2) 1]
Por lo tanto:
Integral original = ∫ (0, ∞){[2(T2) 1]-[(T2) 1]}/[2(T2 ) 1 ][(T2)
=∫(0, ∞)[2(t^2) 1]/[2(t^2) 1][(t^2) 1]dt- p>
∫(0, ∞)[(t^2) 1]/[2(t^2) 1][(t^2) 1]dt
=∫( 0, ∞)1/[(t^2) 1]dt-∫(0, ∞) 1/[2(t^2) 1]dt
=Arctangente|(0, ∞) - (1 /√2)∫(0, ∞)1/[(√2t)^2 1]d(√2t)
=[Arctangente - (1/√2)Arctangente√ 2t ]|(0 , ∞)
=[1-(1/√2)]π