Integration Involving Square Root

[cse63146]

Homework Statement

$$\int \sqrt{t^8 + t^6 } dt$$

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.

 

[rock.freak667]

write t⁸+t⁶ as t⁶(t²+1) and then try a trig substitution.

 

[tiny-tim]

$$\int \sqrt{t^8 + t^6 } dt$$
…
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:

$$\int t^3\,\sqrt{1 + t^2} dt$$

EDIT: waaa! rock.freak667 beat me to it!

 

[cse63146]

where $$t = tan\vartheta \ and \ dt =sec^2 \vartheta \ d\vartheta$$?

 

[Defennder]

Yes, looks ok as an approach.

 

[cse63146]

stuck again (that was fast)

so I got it to this: $$\int tan^6 \vartheta (sec^2 \vartheta )$$

do I use integration by parts here?

or would I use a second substitution with u = tan and du = sec^2 ?

 

[Defennder]

I was thinking about this a bit more and I realize you don't have to do trigo substitution. You have $$\int t^3 \sqrt{t^2+1}dt$$. Do it by parts here, multiple times.

 

[tiny-tim]

Hi cse63146!

stuck again (that was fast)

so I got it to this: $$\int tan^6 \vartheta (sec^2 \vartheta )$$

Nooo … it's $$\int tan^3 \vartheta (sec^2 \vartheta )$$ …

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.

 

[cse63146]

so integration by parts is the following:

$$\int f(x)'g(x) \ dx = f(x)g(x) - \int f(x)g'(x) \ dx$$

I chose f'(x) to be $$\sqrt{1 + t^2}$$ 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.

 

[tiny-tim]

so integration by parts is the following:

$$\int f(x)'g(x) \ dx = f(x)g(x) - \int f(x)g'(x) \ dx$$

I chose f'(x) to be $$\sqrt{1 + t^2}$$ 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 $$t\,\sqrt{1 + t^2}$$

… 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 = … ?

 

[Gib Z]

The easiest method here would probably be the appropriate hyperbolic trig substitution.

 

[cse63146]

Either do integration by parts, staying with t, or (possibly slightly easier) just substitute v = 1 + t², dv = … ?

dv = 2t

so would it look like this:

$$2\int t^2 \sqrt{v} dv$$

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.

 

[tiny-tim]

dv = 2t

so would it look like this:

$$2\int t^2 \sqrt{v} dv$$

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.

$$2\int t^2 \sqrt{v} dv$$ isn't quite finished, is it?

you can make it $$2\int (v - 1) \sqrt{v} dv$$ or just $$2\int (v^{3/2}\,-\,v^{1/2}) dv$$

(when you make a substitution, you must change everything! )

 

[cse63146]

Ah, that makes sense. Once I do the integration, I just "return" t back into the equation. Thanks.