Calculus II Trig Sub

  • Thread starter pumpkin88
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  • #1
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I'm very unsure on how to do this equation...
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
 

Answers and Replies

  • #2
459
1
sqrt of x^2? you sure you wrote that right?
 
  • #3
459
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you wouldn't substitute in a secant, a tangent would do much better, when you sub in you have to figure out what you should put in the place of that x^3 too!
 
  • #4
5
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yeah. its the integral of x^3 times the sqrt of x^2 + 9
 
  • #5
5
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so would it be 3tan^3 theta times the sq rt of 3tan^2theta +9?
 
  • #6
459
1
and you need d(theta) in there too, plus remember you have to raise the entire term to the power
 
  • #7
5
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would d(theta) be 3sec(theta)tan(theta) ?
 
  • #8
so would it be 3tan^3 theta times the sq rt of 3tan^2theta +9?
Remember your letting x = 3tan (theta).

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

what is the derivative in terms of d (theta) ?
 
  • #10
5
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is this correct so far?

9 [tex]\int[/tex] tan[tex]^{3}[/tex] [tex]\theta[/tex] [tex]\sqrt{[/tex] sec [tex]^{2}[/tex] [tex]\theta[/tex]
 
  • #11
is this correct so far?

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

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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