- #1
- 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.
Hi cse63146![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.
:rofl:stuck again (that was fast)
Nooo … it's [tex]\int tan^3 \vartheta (sec^2 \vartheta )[/tex] …so I got it to this: [tex]\int tan^6 \vartheta (sec^2 \vartheta )[/tex]
Easiest is probably to go to u straight from t (without going through θ), and then use integration by parts.do I use integration by parts here?
or would I use a second substitution with u = tan and du = sec^2 ?
No … always mix-and-match to get an easy f'(x) …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 … as Defennder says, you don't need a trig sub!… so I need to do a trig sub …
dv = 2tEither do integration by parts, staying with t, or (possibly slightly easier) just substitute v = 1 + t^{2}, dv = … ?
Hi cse63146! I hope you feel completely better soon.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.