1.∫sec^3/2xtanxdx= 2.∫sin2xdx/根号(1+cos^2x)=

1个回答

  • ∫ (secx)^(3/2) • tanx dx

    = ∫ (secx)^(1 + 1/2) • tanx dx

    = ∫ √(secx) • secxtanx dx

    = ∫ √(secx) d(secx)

    = (2/3)(secx)^(3/2) + C

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    ∫ (sin2x)/√(1 + cos²x) dx = ∫ (2sinxcosx)/√(1 + cos²x) dx

    (v = cos²x,dv = (2cosx)(- sinx) dx)

    = ∫ - 1/√(1 + v) dv

    = - ∫ (1 + v)^(- ½) d(1 + v)

    = - [(1 + v)^(- ½ + 1)]/(- ½ + 1) + C

    = - 2√(1 + v) + C

    = - 2√(1 + cos²x) + C