Calculus
posted by Salman .
For the following integral find an appropriate TRIGNOMETRIC SUBSTITUTION of the form x=f(t) to simplify the integral.
INT((4x^23)^1.5) dx x=?

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.