求tan^4(x) *sec(x) 的不定积分

1个回答

  • 公式:

    J_(n) = ∫ sec^n(x) dx

    J_(n) = [sec^(n - 1)(x) sin(x)]/(n - 1) + (n - 2)/(n - 1) J_(n - 2)

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    ∫ tan^4(x) secx dx

    = ∫ (sec²x - 1)² secx dx

    = ∫ (sec^4(x) - 2sec²x + 1) secx dx

    = ∫ sec^5(x) dx - 2∫ sec³x dx + ∫ secx dx

    = J_(5) - 2J_(3) + J

    = (1/4) sec^4(x) sinx + 3/4 J_(3)

    = (1/4) sec^4(x) sinx - (5/4)J_(3) + J

    = (1/4) sec^4(x) sinx - (5/4)[(1/2) sec²x sinx + 1/2 J] + J

    = (1/4) sec^4(x) sinx - (5/8) sec²x sinx + (3/8)J

    = (1/4) sec^4(x) sinx - (5/8) sec²x sinx + (3/8)ln|secx + tanx| + C