Calculus 2 Integral

  • Thread starter waterchan
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  • #1
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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 )
 

Answers and Replies

  • #2
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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
dextercioby
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You could try a hyperbolic substitution.

[tex] \frac{x}{2\sqrt{2}}=\cosh t [/tex]

Daniel.
 
  • #4
Or just let u = x2 - 8, so x^3dx = (u+8)du/2.

--J
 
  • #5
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Or you can do trigonometric substitution.

[tex] x = \sqrt{8} sec\theta; [/tex]

[tex] dx = \sqrt{8} sec\theta tan\theta d\theta ;[/tex]

and [tex] x^3 = 8\sqrt{8} sec^3\theta [/tex]

Edit message:

Blah i would definitely go for Justin's substitutions.
 
Last edited:
  • #6
GCT
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Trig substitution will definitely work, you can simplify towards the step

[tex]64 \sqrt{(8)} \int sec^{4} \theta~tan^{2} \thetad \theta [/tex]

[tex]64 \sqrt{(8)} \int sec^{2} \theta~(1+tan^{2} \theta )tan^{2} \theta d \theta [/tex]

[tex]u=tan \theta [/tex]

[tex]du= sec^{2} \theta d \theta [/tex]
 

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