
Junior Member
Evaluate the integral of t^5/(sqrt(t^2 + 2))dt
Hello,
This problem brings me to tears. I have spent an hour on it, about half of that solving it and the other half trying to figure out why i'm getting the wrong answer.
And I still have 15 more of these to do for homework...
Find the integral of: t^5/(sqrt(t^2 + 2)) dt
I need to do it with trigonometric substitution.
t = sqrt(2)tan(z)
dt = sqrt(2)sec^2(z)dz
I will skip some steps since it's so long...
integral of: [5 * sqrt(2) * tan^5(z)sec^2(z)dz]/[sqrt(2tan^2(z) + 2)]
= 5 * the integral of: tan^5(z)sec(z)dz
...
= 5 * integral of: (sec^2(z)  1)^2 tan(z)sec(z)dz
Let u = sec(z)
du = tan(z)sec(z)dz
5 * integral of: (u^2  1)^2du
= 5 * [1/5 * u^5 + u  2/3 * u^3] + C ... substitute secant for u
evaluated right triangle using tan(z) = t/sqrt(2), with secant being = sqrt(2 + t^2)/sqrt(2) ... numerator is hypotenuse, sqrt(2) is side adjacent to z
answer: 5[(1/5 * [sqrt(2 + t^2)/sqrt(2)]^5 + sqrt(2 + t^2)/sqrt(2)  2/3 * [sqrt(2 + t^2)/sqrt(2)]^3] + C
which is apparently not correct. this is a horrifically long problem that I wouldn't wish on anyone but does anyone have some ideas during which part I am making a mistake?
Thanks

New Member
Re: Evaluate the integral of t^5/(sqrt(t^2 + 2))dt
Don't cry my friend, try this
After 5 * the integral of: tan^5(z)sec(z)dz (where I think 5 should be (2)^(5/2))
write
= 5* the integral of:tan^4(z) tan(z) sec(z) dz
= 5* the integral of:{tan^2(z)}^2 tan(z) sec(z) dz
= 5* the integral of:{sec^2(z)  1}^2 tan(z) sec(z) dz
= 5* the integral of:{sec^4(z)+12sec^2(z)} tan(z) sec(z) dz
Now let B = sec(z)
then B' = sec(z) tan(z) dz
Rest should be easy for an intellegent person like you
Regards

Elite Member
Re: Evaluate the integral of t^5/(sqrt(t^2 + 2))dt
With all due respect, you really should learn LaTex. Folks don't want to look through typed out format to look for a mistake. It's easier to spot if it's in math type font.
Anyway. [tex]t=\sqrt{2}tan{\theta}, \;\ dt=\sqrt{2}sec^{2}{\theta}d{\theta}[/tex]
When you make the subs you get:
[tex]4\sqrt{2}\int{tan^{5}{\theta}sec{\theta}}d{\theta}[/tex]
You can integrate by using the reduction formula for sec and tan or do it the long way:
[tex]\int{tan^{5}{\theta}sec{\theta}}d{\theta}=\int(sec ^{2}{\theta}1)^{2}sec{\theta}tan{\theta}d{\theta}=[/tex]
[tex]\int(sec^{4}{\theta}2sec^{2}{\theta}+1)sec{\theta}tan{\theta}d{\theta}[/tex]
[tex]=\frac{1}{5}sec^{5}{\theta}\frac{2}{3}sec^{3}{\theta}+sec{\theta}[/tex]
Now, resub [tex]{\theta}=tan^{1}(\frac{t}{\sqrt{2}})[/tex]
I know, this is booger of a problem to do by hand. Let it be known what you get after resubbing. Don't forget the
[tex]4\sqrt{2}[/tex] hanging up there. I mostly just pick it up at the end instead of carrying it along for the ride.

Junior Member
Re: Evaluate the integral of t^5/(sqrt(t^2 + 2))dt
Hello,
Thanks for the reply. You have helped me isolate where the problem is. The problem is with the number I pulled out of the integral at the beginning. It should be 4sqrt(2) instead of 5. Everything else is ok  once I change that and pull it down to the final expression I listed to replace the 5, I get the right numerical answer. What a mess this simple little math mistake caused!

Elite Member
Re: Evaluate the integral of t^5/(sqrt(t^2 + 2))dt
Did you get:
[tex]\frac{\sqrt{t^{2}+2}(3t^{4}8t^{2}+32)}{15}[/tex]
Or some variation thereof?.
Mind you, you may have an equivalent solution only not in that form.
Sometimes it's difficult to see.

Junior Member
Re: Evaluate the integral of t^5/(sqrt(t^2 + 2))dt
The answer I ended up with was
answer: 4sqrt(2) * [(1/5 * [sqrt(2 + t^2)/sqrt(2)]^5 + sqrt(2 + t^2)/sqrt(2)  2/3 * [sqrt(2 + t^2)/sqrt(2)]^3] + C
which is different than the book said, but I picked a '3' and evaluated it in both answers for 't' and got the same result so I assume they are equal

Junior Member
Re: Evaluate the integral of t^5/(sqrt(t^2 + 2))dt
Find the integral of: t^5/(sqrt(t^2 + 2)) dt
I need to do it with trigonometric substitution.
>> Do you? How about a nice rationalizing subst:
u = sqrt(t^2 + 2)
u^2 = t^2 + 2
t^2 = u^2  2
2t dt = 2u dt
t dt = u dt
t^5/(sqrt(t^2 + 2)) dt =
t^4/(sqrt(t^2 + 2)) t dt =
(u^2  2)^2/u u du =
(u^2  2)^2 du = from here, you're on your own.
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