Find the Integral of x^3 * (x^2 - 8)^1/2 | Calculus 2

[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 )

 

[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.

 

[dextercioby]

You could try a hyperbolic substitution.

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

Daniel.

 

[Justin Lazear]

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

--J

 

[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.

 

[GCT]

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

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

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

$$u=tan \theta$$

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