Results 1 to 7 of 7

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

  1. #1
    Junior Member
    Join Date
    Sep 2007
    Posts
    243

    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. #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. #3
    Elite Member
    Join Date
    Sep 2005
    Posts
    7,293

    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.

  4. #4
    Junior Member
    Join Date
    Sep 2007
    Posts
    243

    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. #5
    Elite Member
    Join Date
    Sep 2005
    Posts
    7,293

    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.

  6. #6
    Junior Member
    Join Date
    Sep 2007
    Posts
    243

    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. #7
    Junior Member
    Join Date
    Dec 2007
    Posts
    124

    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.

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •