? asked in Science & MathematicsMathematics · 6 years ago

Integration by U substitution pls help?

What's the

1-integration of U sqrt 1-U^2 du and

2-integration of sqrt X^3-3/sqrt X^11 dx

3 Answers

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  • cidyah
    Lv 7
    6 years ago
    Best Answer

    1) ∫ u sqrt(1-u^2) du

    Let t=1-u^2

    dt = -2udu

    u du = (-1/2) dt

    ∫ u sqrt(1-u^2) du = (-1/2) ∫ sqrt(t) dt = (-1/2) ∫ t^(1/2) dt = (-1/2) t^(1/2+1)/(1/2+1) = (-1/2)(2/3) t^(3/2)

    = (-1/3) t^(3/2)

    = (-1/3) (1-u^2)^(3/2) + C

    2) ∫ sqrt(x^3-3) / sqrt(x^11) dx

    = ∫ sqrt(x^3-3) / x^(11/2) dx

    Factor out x^3 from the square root

    = ∫ x^(3/2) sqrt(1-3/x^3) / x^(11/2) dx

    = ∫ sqrt(1-3/x^3) / x^(8/2) dx

    = ∫ sqrt(1-3/x^3) / x^4 dx

    Let u = 1-3/x^3

    u=-3 x^(-3)

    du=(-3)(-3) x^(-4)

    du = 9/x^4 dx

    1/x^4 dx = (1/9) du

    ∫ sqrt(1-3/x^3) / x^4 dx = (1/9) ∫ sqrt(u) du

    = (1/9) ∫ u^(1/2) du

    = (1/9) u^(1/2+1)/(1/2+1)

    = (1/9)(2/3)u^(3/2)

    =(2/27) u^(3/2)

    = (2/27) (1-3/x^3)^(3/2) + C

  • Como
    Lv 7
    6 years ago

    1.

    I = ∫ x (1 - x²)^(1/2) dx

    Let u = 1 - x²

    du = -2x dx

    -du/2 = x dx

    I = (-1/2) ∫ u^(1/2) du

    I = (-1/3) (2/3) u^(3/2) + C

    I = (-2/9) (1 - x²) + C

    2.

    As given MUST be read as x³ - [ 3/x^(11) ] so will answer accordingly.

    I = ∫ x³ - 3 x^(11/2) dx

    I = x^4 / 4 - (6/13) x^(13/2) + C

  • 6 years ago

    1) 2[(1-u^2)^(3/2)]3 + c. Use appropriate formula.

    2) Can't understand sqrt(x^3-3) or sqrt(x^3) - 3......

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