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January 30, 2015

January 30, 2015

Posted by **Salman** on Thursday, December 10, 2009 at 4:18pm.

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

- Calculus -
**Reiny**, Thursday, December 10, 2009 at 5:12pmI 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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