Integration Involving Square Root

In summary, the conversation discusses how to solve the integral \int \sqrt{t^8 + t^6 } dt by rewriting it as \int t^3\,\sqrt{1 + t^2} dt and using integration by parts or a substitution with v = 1 + t^2. It is also mentioned that a hyperbolic trig substitution could be used.
  • #1
cse63146
452
0

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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  • #2
write t8+t6 as t6(t2+1) and then try a trig substitution.
 
  • #3
cse63146 said:
[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
where [tex]t = tan\vartheta \ and \ dt =sec^2 \vartheta \ d\vartheta [/tex]?
 
  • #5
Yes, looks ok as an approach.
 
  • #6
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
I was thinking about this a bit more and I realize 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
Hi cse63146! :smile:
cse63146 said:
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
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
cse63146 said:
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
The easiest method here would probably be the appropriate hyperbolic trig substitution.
 
  • #12
tiny-tim said:
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
cse63146 said:
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
Ah, that makes sense. Once I do the integration, I just "return" t back into the equation. Thanks.
 

1. What is integration involving square root?

Integration involving square root is a technique used in calculus to find the antiderivative of a function that involves a square root. It is a type of indefinite integral and is used to find the area under a curve.

2. How is integration involving square root different from regular integration?

The main difference between integration involving square root and regular integration is that the former involves functions with a square root, while the latter can involve any type of function. Integration involving square root requires a different set of rules and techniques to solve compared to regular integration.

3. What are the steps to solve an integration involving square root problem?

The steps to solve an integration involving square root problem are as follows:

  1. Simplify the function by factoring out any constants or variables.
  2. Use substitution to rewrite the function in terms of a new variable.
  3. Apply the power rule to the new function, if possible.
  4. Integrate the new function using the power rule or other integration rules.
  5. Substitute the original variable back into the solution.

4. When is integration involving square root used in real life?

Integration involving square root is used in various real-life applications, such as calculating the work done by a variable force, finding the center of mass of an object with varying density, and determining the velocity of an object under the influence of gravity. It is also used in engineering, physics, and economics.

5. Are there any common mistakes to avoid when solving integration involving square root problems?

One common mistake to avoid when solving integration involving square root problems is forgetting to use substitution and trying to apply regular integration rules. Another mistake is not simplifying the function before attempting to integrate it. It is also important to be careful with signs and constants when using the power rule. It is always recommended to double-check the final solution by differentiating it to ensure it is the correct antiderivative of the original function.

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