- #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.

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- Thread starter cse63146
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- #1

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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.

- #2

rock.freak667

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write t^{8}+t^{6} as t^{6}(t^{2}+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!

First thing would be to rewrite as:

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

EDIT: waaa!

- #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 ?

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 ?

Last edited:

- #7

Defennder

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- #8

tiny-tim

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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

- #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.

[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

in this case choose f'(x) to be [tex]t\,\sqrt{1 + t^2}[/tex]

… so I need to do a trig sub …

No … as

Either do integration by parts, staying with t,

- #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 + t^{2}, dv = … ?

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.

[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]

(when you make a substitution, you must change

- #14

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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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