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Math (Cal)

posted by on .

I having difficulty integrating this
Ingrate (x^3)/sqrt(x^2 - 9)

I know that x= 3sec(theta)
Sqrt(x^2-9)= 3tan(theta)
x^3 = 27Sec^3(theta)

Once I put that all together I'm not sure what to do next.

  • Math (Cal) - ,

    so far, so good. But you forgot about dx:

    dx = 3secθtanθ dθ

    So, now you have

    ∫27sec^3(θ) / 3tanθ * 3secθtanθ
    = 27∫sec^4(θ) dθ
    = 27∫(tan^2(θ)+1) sec^2(θ) dθ
    Now if you let u = tanθ, you have
    = 27∫(u^2+1) du
    = 27(u^3/3 + u)
    = 27(tan^3(θ)/3 + tanθ)
    = 9tanθ (tan^2(θ) + 3)

    Now, the original substitutions,
    tanθ = 1/3 √(x^2-9), so we wind up with

    3√(x^2-9) (1/9 (x^2-9) + 3)
    = 1/3 √(x^2-9) (x^2-9+27)
    = 1/3 √(x^2-9) (x^2+18)

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