- #1

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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 )

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- Thread starter waterchan
- Start date

- #1

- 23

- 0

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

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- #3

- 13,141

- 696

You could try a hyperbolic substitution.

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

Daniel.

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

Daniel.

- #4

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

--J

--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.

[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

Science Advisor

Homework Helper

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[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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