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March 4, 2015

March 4, 2015

Posted by **Salman** on Thursday, December 10, 2009 at 10:31am.

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

- Calculus -
**drwls**, Thursday, December 10, 2009 at 11:40amu = 4x^2 -3

(1/8)du = x dx

(1/8)INT u^(3/2) du

(1/8)(2/5)u^(5/2)

(1/20)(4x^2 -3)^(5/2)

I don't see why they demand a trigonometric solution. This seems like the logical one to use.

- Calculus -
**Anonymous**, Thursday, December 10, 2009 at 4:07pmNo it is required to use a trignometric substitution. x cannot be a variable in the final answer

- Calculus -
**Bun**, Thursday, December 10, 2009 at 5:51pmyou can do like this:

(4x^2-3)^1.5 = (4x^2-3).sq(4x^2-3)dx

Let 2x = Sq3.sec(det)

so, 4x^2 -3 = 3.sec^2(det)-3

= 3.tan^2(det)

new INT= INt(3^1.5tan^3(det)d(det)

Use Pythagore to solve the relation between x and the angle det.

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