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Integration problem

  1. Feb 20, 2010 #1
    integral from 0 to 2 of (1+9x^4)^1/2 dx

    I was thinking that I could use a trig substitution to solve for it so it would be:
    [2]\int[/0] [9(1/9 +x^4)]^1/2
    so [2]\int[/0]3[(1/9+x^4)]^1/2

    my problem is that I can understand that my a would b 1/3, and that it should be atantheta, but what I can't seem to get is that it is an X^4. I understand that it is (x^2)^2, but again, how do I do it because I only know how to deal with it when it is x^2. If anyone has any suggestions how to deal with this problem, that would be great.
     
  2. jcsd
  3. Feb 21, 2010 #2
    ok here goes nothing
    i start out by x=1/(sqrt(3))tan(u) dx=1/sqrt(3)sec(u)^2
    so then w have 1/(sqrt3)*sqrt(1+tan(u)^4)*sec(u)^2du
    then i am using the formula for reducing powers for tan(u)^2
    which is tan(u)^2=(1-cos(2u))/(1+cos(2u)) then substituting
    this in for tan^2 and then foiling this
    and then getting a common denominator with the one inside the radical and combining like terms to give me
    sqrt(2+2cos(2x)/(1+2cos(2x)+cos(2x)^2)
    then i factor the top and bottom
    can u see how it factors 2(1+u)/(u+1)^2 u=cos(2x)
    leaving us with 2/(cos(2x)+1) which if you notice this is the inverse of the formula for reducing powers for cos^2) so this is sec^2(u) so this simplifies beautifully to
    sqrt(sec^2)(sec^2)1/(sqrt(3)
    which becomes sec^3(u) so we then integrate sec^3 which we know how to do
    which becomes 1/2(secu)tan(u)+1/2ln|secu+tanu| and our 1/sqrt(3) out front
    then from our original substitution x=1/sqrt(3)tanu we simply draw our little triangle
    and find the sec and tan of it and back substitute . i hope this is right ,
     
    Last edited: Feb 21, 2010
  4. Feb 21, 2010 #3
    wait...how did you get a cube root?
     
  5. Feb 21, 2010 #4

    Mark44

    Staff: Mentor

    It's not a cube root -- sqrt(3) is the square root of 3. I think that's what you're referring to.
     
  6. Feb 21, 2010 #5
    I am still lost though as to how you got that. And why is it 1 over that?
     
  7. Feb 21, 2010 #6

    Mark44

    Staff: Mentor

    I think what cragar did was to use a trig substitution. I'll let him/her jump back in and correct me if the details aren't right.

    Draw a right triangle with one acute angle labelled u. The opposite side is 3x^2 and the adjacent side is 1. The hypotenuse is sqrt(1 + 9x^4). From this triangle, tan u = 3x^2, so x = sqrt(tan u)/sqrt(3). Also, sec^2(u)du = 6x dx. With these relationships you can rewrite the original integral so that it is in terms of u and du.
     
  8. Feb 21, 2010 #7
    ok, that makes sense but how do you plug it back it? Do you plug it back in so that the 3x^2 is substituted by tanu, or is just the x substituted by tan^(1/2)/root3? Thats where I understood that it should have been a trig sub, but I didn't know how to do it with a x^4
     
  9. Feb 21, 2010 #8

    Mark44

    Staff: Mentor

    You substitute based on the the relations I gave in post #6, that are based on the right triangle I described. You can get everything from that.
     
  10. Feb 21, 2010 #9
    oh, ok. I get it now, Thank you
     
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