∫(0,1)根号下[1-x^2]dx=
1个回答

令x=sint ,dx=costdt 当x=0时,t=0,当x=1时,t=π/2

∫(0,1)根号下[1-x^2]dx

= ∫(0,π/2)√(1-sin²t)costdt

= ∫(0,π/2)√(1-sin²t)costdt

=∫(0,π/2)cos²tdt

=∫(0,π/2) [(cos2t +1)/2]dt

=1/2∫(0,π/2)cos2t+1dt

=1/2[1/2∫(0,π/2)cos2td2t+∫(0,π/2)dt]

=1/2[1/2(sinπ-sin0)+(π/2-0)]

=1/2×π/2

=π/4