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For the following integral find an appropriate TRIGNOMETRIC SUBSTITUTION of the form x=f(t) to simplify the integral.

INT((4x^2-3)^1.5) dx x=?

  • Calculus -

    I would try something like this:

    Integral (4x^2 - 3)dx
    = integral (3(4x^2/3 - 1)dx

    let (4/3)x^2 = sin^2Ѳ
    then (2/√3)x = sinѲ
    x = (√3/2)sinѲ
    dx/dѲ = (√3/2)cosѲ
    dx = (√3/2)cosѲ dѲ

    so your integral turns into

    integral (3(sin^2Ѳ - 1)^1.5 (√3/2)cosѲ dѲ

    However, I see a problem arising.
    sin^2 Ѳ - 1 ≤ 0 because of the sine function and we cannot raise a negative value to an exponent of 1.5

    Perhaps you can see how to get around that.

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