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

1. ## 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

2. ## 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)+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

3. ## 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. $t=\sqrt{2}tan{\theta}, \;\ dt=\sqrt{2}sec^{2}{\theta}d{\theta}$

When you make the subs you get:

$4\sqrt{2}\int{tan^{5}{\theta}sec{\theta}}d{\theta}$

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

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

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

$=\frac{1}{5}sec^{5}{\theta}-\frac{2}{3}sec^{3}{\theta}+sec{\theta}$

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

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

$4\sqrt{2}$ hanging up there. I mostly just pick it up at the end instead of carrying it along for the ride.

4. ## 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!

5. ## Re: Evaluate the integral of t^5/(sqrt(t^2 + 2))dt

Did you get:

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

Or some variation thereof?.

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

Sometimes it's difficult to see.

6. ## 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

7. ## 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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