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Another Math Question

  1. Nov 7, 2003 #1
    This question is ugly,


    His hints were first use the substitution u=tan(θ) then use the substitution u=x^2

    I then had 2∫(x^2dx)/(sec^2(θ)

    I let x=tan(θ) so that dx=sec^2(θ)dθ

    And then got 2∫tan^2(θ)dθ

    which I got as 2tan(θ)-2θ+C, and d/dx of that is not √tan(θ)

    Any thoughts?
  2. jcsd
  3. Nov 7, 2003 #2


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    It looks like you tried to do too much at once and got your dummy variables confused!

    You need to complete all of the work for one substitution before you can move onto the next substitution. In particular, after making the substitution u = tan θ, you have to rewrite all occurences of θ in terms of u.

    (hint: first write (sec θ)^2 in terms of tan θ)
  4. Nov 8, 2003 #3
    ok I did that and ended up with

    2∫(x^2dx)/(1+x^4) and from there I am kinda stuck on what I want to do, should I let x^2=tan(θ)? But I think if I do that I end up with a 2x that I dont want...
  5. Nov 8, 2003 #4


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    Hrm. Are you permitted to use complex numbers? The denominator of that is easy to factor over the complex domain, allowing you to use partial fractions.
  6. Nov 8, 2003 #5
    hmmm complex numbers, not sure lol. This was in the chapter of integration techniques. Integration by parts, Substitution, Trig Substitution, Partial Fraction Decomposition, Improper Integrals and Hyperbolics. Not sure exactly what complex numbers are or if we can use them.


    (x^4+1)= (x^2-√(2)x+1)(x^2+√(2)x+1) and then use partial fraction decomp?... lol maybe thats why he told us that factor
    Last edited: Nov 8, 2003
  7. Nov 8, 2003 #6


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    Yes, in terms of complex numbers, x4+1 factors as

    Of course, the product of the first two of those is
    x2-&radic(2)x+ 1 and the product of the last two is
    x2+&radic(2)x+ 1 just as you have.
  8. Nov 8, 2003 #7
    well hours later I finally solved it, and I came out with an insanely ugly answer. After I did the partial fraction decomp I completed the square twice and used some trig substitutions followed by another substitution.

    this is what I ended up with, not sure if it will be easier to type it, or to scan it lol

    (1/8)[ ln|(√(2))/2) / ((√tan(θ))-(√(2))/2)^2+(1/2))| - tan^-1((√tan(θ)-(√(2))/2) / ((√(2))/2)) - ln|(√(2))/2) / ((√tan(θ))+(√(2))/2)^2+(1/2))| - tan^-1((√tan(θ)+(√(2))/2) / ((√(2))/2)) ] + C

    ughh is that ugly, I have no clue if it is right or not, but I am not gonna even bother with it anymore [zz)]

    which is pretty darn close to what the integrator said, after some moving around of course.
    Last edited: Nov 8, 2003
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