Evaluate the integral of t^5/(sqrt(t^2 + 2))dt

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MarkSA

Junior Member
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Sep 8, 2007
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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
 
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)+1-2sec^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
 
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. \(\displaystyle t=\sqrt{2}tan{\theta}, \;\ dt=\sqrt{2}sec^{2}{\theta}d{\theta}\)

When you make the subs you get:

\(\displaystyle 4\sqrt{2}\int{tan^{5}{\theta}sec{\theta}}d{\theta}\)

You can integrate by using the reduction formula for sec and tan or do it the long way:

\(\displaystyle \int{tan^{5}{\theta}sec{\theta}}d{\theta}=\int(sec^{2}{\theta}-1)^{2}sec{\theta}tan{\theta}d{\theta}=\)

\(\displaystyle \int(sec^{4}{\theta}-2sec^{2}{\theta}+1)sec{\theta}tan{\theta}d{\theta}\)

\(\displaystyle =\frac{1}{5}sec^{5}{\theta}-\frac{2}{3}sec^{3}{\theta}+sec{\theta}\)

Now, resub \(\displaystyle {\theta}=tan^{-1}(\frac{t}{\sqrt{2}})\)

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

\(\displaystyle 4\sqrt{2}\) hanging up there. I mostly just pick it up at the end instead of carrying it along for the ride.
 
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!
 
Did you get:

\(\displaystyle \frac{\sqrt{t^{2}+2}(3t^{4}-8t^{2}+32)}{15}\)

Or some variation thereof?.

Mind you, you may have an equivalent solution only not in that form.

Sometimes it's difficult to see.
 
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
 
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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