- #1

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Someone PLEASE HELP!!!

[tex]\int[/tex]x[tex]^{3}[/tex][tex]\sqrt{}x^{2}+9[/tex]

i do know that

if you...

let x = 3sec [tex]\vartheta[/tex]

dx = 3 sec [tex]\vartheta[/tex] tan [tex]\vartheta[/tex] du

- Thread starter pumpkin88
- Start date

- #1

- 5

- 0

Someone PLEASE HELP!!!

[tex]\int[/tex]x[tex]^{3}[/tex][tex]\sqrt{}x^{2}+9[/tex]

i do know that

if you...

let x = 3sec [tex]\vartheta[/tex]

dx = 3 sec [tex]\vartheta[/tex] tan [tex]\vartheta[/tex] du

- #2

- 459

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sqrt of x^2? you sure you wrote that right?

- #3

- 459

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- #4

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yeah. its the integral of x^3 times the sqrt of x^2 + 9

- #5

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so would it be 3tan^3 theta times the sq rt of 3tan^2theta +9?

- #6

- 459

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and you need d(theta) in there too, plus remember you have to raise the entire term to the power

- #7

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would d(theta) be 3sec(theta)tan(theta) ?

- #8

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Remember your letting x = 3tan (theta).so would it be 3tan^3 theta times the sq rt of 3tan^2theta +9?

x^3 means 9tan^3 (theta) and the other term becomes sqrt [ 9tan^2 (theta) + 9 ]

Factor out a square root of 9 out of the square root and use the identity: 1 + tan^2 x = sec^2 x to "simplify" your problem.

You should be left with an integration of a sec to the third and a tangent to the third.

- #9

- 133

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if you let x = 3tan (theta)would d(theta) be 3sec(theta)tan(theta) ?

what is the derivative in terms of d (theta) ?

- #10

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9 [tex]\int[/tex] tan[tex]^{3}[/tex] [tex]\theta[/tex] [tex]\sqrt{[/tex] sec [tex]^{2}[/tex] [tex]\theta[/tex]

- #11

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Close.

9 [tex]\int[/tex] tan[tex]^{3}[/tex] [tex]\theta[/tex] [tex]\sqrt{[/tex] sec [tex]^{2}[/tex] [tex]\theta[/tex]

Like lumbar said, you forgot the d(Theta).

if you let x = 3tan(Theta)

derive the above equation in terms of d(Theta)

you should get:

dx = <derivative of right hand equation>

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