Calculus 2 Integral

1. May 18, 2005

waterchan

How would you find the integral of x^3 * ( x^2 - 8 )^1/2 ?

I've tried everything I can think of; substitution / trigonometric substitution, partial fractions, integration by parts, but I just can't get the answer. The answer is supposed to be:

1/15(x^2-8)^3/2 * (3x^2 + 16 )

2. May 18, 2005

whozum

Separate x^3 into x^2*x, then do integration by parts with x^2 for one and x(x^2-8)^.5 for the other.

3. May 18, 2005

dextercioby

You could try a hyperbolic substitution.

$$\frac{x}{2\sqrt{2}}=\cosh t$$

Daniel.

4. May 19, 2005

Justin Lazear

Or just let u = x2 - 8, so x^3dx = (u+8)du/2.

--J

5. May 19, 2005

zeronem

Or you can do trigonometric substitution.

$$x = \sqrt{8} sec\theta;$$

$$dx = \sqrt{8} sec\theta tan\theta d\theta ;$$

and $$x^3 = 8\sqrt{8} sec^3\theta$$

Edit message:

Blah i would definitely go for Justin's substitutions.

Last edited: May 19, 2005
6. May 19, 2005

GCT

Trig substitution will definitely work, you can simplify towards the step

$$64 \sqrt{(8)} \int sec^{4} \theta~tan^{2} \thetad \theta$$

$$64 \sqrt{(8)} \int sec^{2} \theta~(1+tan^{2} \theta )tan^{2} \theta d \theta$$

$$u=tan \theta$$

$$du= sec^{2} \theta d \theta$$