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Integration Involving Square Root

  • Thread starter cse63146
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  • #1
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Homework Statement



[tex]\int \sqrt{t^8 + t^6 } dt[/tex]

Homework Equations





The Attempt at a Solution



I'm not sure what to do next, can someone point me in the right direction? Thank you.
 

Answers and Replies

  • #2
rock.freak667
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write t8+t6 as t6(t2+1) and then try a trig substitution.
 
  • #3
tiny-tim
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[tex]\int \sqrt{t^8 + t^6 } dt[/tex]

I'm not sure what to do next, can someone point me in the right direction? Thank you.
Hi cse63146! :smile:

First thing would be to rewrite as:

[tex]\int t^3\,\sqrt{1 + t^2} dt[/tex] :wink:

EDIT: waaa! rock.freak667 beat me to it! :cry:
 
  • #4
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where [tex]t = tan\vartheta \ and \ dt =sec^2 \vartheta \ d\vartheta [/tex]?
 
  • #5
Defennder
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Yes, looks ok as an approach.
 
  • #6
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stuck again (that was fast)

so I got it to this: [tex]\int tan^6 \vartheta (sec^2 \vartheta )[/tex]

do I use integration by parts here?

or would I use a second substitution with u = tan and du = sec^2 ?
 
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  • #7
Defennder
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I was thinking about this a bit more and I realise you don't have to do trigo substitution. You have [tex]\int t^3 \sqrt{t^2+1}dt [/tex]. Do it by parts here, multiple times.
 
  • #8
tiny-tim
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Hi cse63146! :smile:
stuck again (that was fast)
:rofl:
so I got it to this: [tex]\int tan^6 \vartheta (sec^2 \vartheta )[/tex]
Nooo … it's [tex]\int tan^3 \vartheta (sec^2 \vartheta )[/tex] …
do I use integration by parts here?

or would I use a second substitution with u = tan and du = sec^2 ?
Easiest is probably to go to u straight from t (without going through θ), and then use integration by parts. :wink:
 
  • #9
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so integration by parts is the following:

[tex]\int f(x)'g(x) \ dx = f(x)g(x) - \int f(x)g'(x) \ dx[/tex]

I chose f'(x) to be [tex]\sqrt{1 + t^2}[/tex] so I could eventually get rid of t^3, but I have no idea how to find f(x)

so I need to do a trig sub like this:http://www.freemathhelp.com/forum/viewtopic.php?f=3&t=30001&start=0 but what happens to the square root? It just disappears after he uses the trig identity sin^2 + cos^2 = 1, and it's 25 instead of 5.
 
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  • #10
tiny-tim
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so integration by parts is the following:

[tex]\int f(x)'g(x) \ dx = f(x)g(x) - \int f(x)g'(x) \ dx[/tex]

I chose f'(x) to be [tex]\sqrt{1 + t^2}[/tex] so I could eventually get rid of t^3, but I have no idea how to find f(x)
No … always mix-and-match to get an easy f'(x) …

in this case choose f'(x) to be [tex]t\,\sqrt{1 + t^2}[/tex] :wink:
… so I need to do a trig sub …
No … as Defennder says, you don't need a trig sub!

Either do integration by parts, staying with t, or (possibly slightly easier) just substitute v = 1 + t2, dv = … ? :smile:
 
  • #11
Gib Z
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The easiest method here would probably be the appropriate hyperbolic trig substitution.
 
  • #12
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Either do integration by parts, staying with t, or (possibly slightly easier) just substitute v = 1 + t2, dv = … ? :smile:
dv = 2t

so would it look like this:

[tex]2\int t^2 \sqrt{v} dv[/tex]

would I write dv or dt in this case since I have both t and v.

P.S sorry it took so long to respond, I got sick shortly after this and my fever was gone this morning.
 
  • #13
tiny-tim
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dv = 2t

so would it look like this:

[tex]2\int t^2 \sqrt{v} dv[/tex]

would I write dv or dt in this case since I have both t and v.

P.S sorry it took so long to respond, I got sick shortly after this and my fever was gone this morning.
Hi cse63146! I hope you feel completely better soon. :smile:

[tex]2\int t^2 \sqrt{v} dv[/tex] isn't quite finished, is it?

you can make it [tex]2\int (v - 1) \sqrt{v} dv[/tex] or just [tex]2\int (v^{3/2}\,-\,v^{1/2}) dv[/tex] :wink:

(when you make a substitution, you must change everything! :smile:)
 
  • #14
452
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Ah, that makes sense. Once I do the integration, I just "return" t back into the equation. Thanks.
 

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