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Homework Help: Another tan and sec integral issue

  1. Jun 27, 2006 #1
    Here is another problem that might be related to my last post. These dang sec and tan. Anyway, here goes...

    [tex]\int\frac{x^3}\sqrt{{x^2+1}}\dx[/tex]

    Oh man.. above is a feeble attempt to write with the LaTeX language. that was my first attempt and a terrible one at that. Haha.. Sorry about that. Anyway, to the question.

    I am trying to integrate ∫(x^3)/((x^+1)^(1/2)) dx

    To start, I let x = tan(θ) which lets dx = sec^2(θ) dθ
    Plugging this back into the equation, we get ∫ (tan^3(θ)/((tan^2(θ)+1)^(1/2)) sex^2(θ) dθ.

    Solving the algebra I come to a point that I do not know what to do next.

    That is, ∫ tan^3(θ)sec(θ) dθ.

    Can anyone offer any advice as to what the next step would be?

    Thank you
     
    Last edited: Jun 27, 2006
  2. jcsd
  3. Jun 27, 2006 #2

    arildno

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    Again, don't bother with trig substitutions!
    Let [tex]u(x)=x^{2}, \frac{dv}{dx}=\frac{x}{\sqrt{x^{2}+1}}[/tex]
     
  4. Jun 27, 2006 #3
    Wait, so if I let u = [tex]x^2[/tex], then what does that make the numerator after substituting it back in? Am I missing some fundamental algebra thing here, because I don't see how that works.
     
  5. Jun 27, 2006 #4

    HallsofIvy

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    The way arildno wrote that, it looks like an "integration by parts" that I don't follow! Let u= x2+ 1 so that du= 2x dx or (1/2)du= xdx.
    The numerator, then, is x2(xdx)= (1/2)(u-1)du and the denominator is u1/2.

    Of course, you can do the tan, sec integral: reduce it to sine and cosine:
    [tex]tan^3 \theta sec \theta= \frac{sin^3 \theta}{cos^4 \theta}[/tex]
    Write that as
    [tex]\frac{sin^2 \theta}{cos^4 \theta} sin \theta = \frac{1- cos^2 \theta}{cos^4 \theta} sin \theta[/tex]
    and let u= cos [itex]\theta[/itex].

    (To get the root around the whole thing, use braces: {x^2+ 1}.)
     
  6. Jun 27, 2006 #5

    arildno

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    Don't follow??
    It is perfectly obvious!
    We get [itex]v=\sqrt{x^{2}+1}[/itex], thus:
    [tex]\int\frac{x^{3}}{\sqrt{x^{2}+1}}dx=x^{2}\sqrt{x^{2}+1}-\int{2}x\sqrt{x^{2}+1}dx=x^{2}\sqrt{x^{2}+1}-\frac{2}{3}(x^{2}+1)^{\frac{3}{2}}+C[/tex]
     
    Last edited: Jun 27, 2006
  7. Jun 27, 2006 #6
    Ok: I tried both approaches here and neither of them seem to be getting me to the correct answer. Or, maybe it is the correct answer and I don't know it. The answer that arildno gave looks correct to me, but I do not follow the algebra. I know, as a calculus student I should be able to follow this no trouble, but it seems a little confusing to me. If you do not want to waste your time explaining it I totally understand.

    The solution that HallsofIvy gave me I completely understand, but I don't seem to come out to a correct answer.

    After
    [tex]\frac{sin^2 \theta}{cos^4 \theta} sin \theta = \frac{1- cos^2 \theta}{cos^4 \theta} sin \theta[/tex]

    and then letting u = cos(θ), I then came out to
    [tex]3sec^3 \theta+sec \theta[/tex]

    That solved for θ, so by building a triangle and finding out what θ is I got
    [tex]1+x^2+3(1+x^2)^3\[/tex]

    That brings about another fundametal issue that I might have a problem with. For example, if I evaluate an integral and come to an answer like the above, [tex]3sec^3 \theta+sec \theta[/tex], for the the [tex]3sec^3 \theta[/tex] part, can I find sec(θ) and then just cube it? If I can, then maybe I got the right answer, but if you can't, then I suppose I am way off. My guess is that you can't.
     
  8. Jun 29, 2006 #7

    VietDao29

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    This line is incorrect.
    You should note that:
    [tex]\int x ^ {\alpha} dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C, \ \alpha \neq -1[/tex]
    Now:
    [tex]\int \frac{1 - \cos ^ 2 \theta}{\cos ^ 4 \theta} \sin \theta d\theta = - \int \frac{1 - \cos ^ 2 \theta}{\cos ^ 4 \theta} d(\cos \theta) = - \int \frac{d(\cos \theta)}{\cos ^ 4 \theta} + \int \frac{d(\cos \theta)}{\cos ^ 2 \theta}[/tex]
    [tex]= - \int \cos ^ {-4} \theta d(\cos \theta) + \int \cos ^ {-2} \theta d(\cos \theta) = \frac{1}{3 \cos ^ 3 \theta} - \frac{1}{\cos \theta} + C[/tex]
    Yes, you can convert [tex]\theta[/tex] back to x like this:
    Since you let:
    [tex]x = \tan \theta \Rightarrow x ^ 2 + 1 = \tan ^ 2 \theta + 1 = \sec ^ 2 \theta \Rightarrow \sec \theta = \sqrt{x ^ 2 + 1}[/tex]
    So:
    [tex]\sec \theta = \sqrt{x ^ 2 + 1}[/tex]. Can you change [tex]\theta[/tex] back to x now? :)
    --------------------
    Here's another method (the u-substitution one)
    Now we must find [tex]\int \frac{x ^ 3 dx}{\sqrt{x ^ 2 + 1}}[/tex], we have:
    [tex]\int \frac{x ^ 3 dx}{\sqrt{x ^ 2 + 1}} = \frac{1}{2} \int \frac{x ^ 2 d(x ^ 2 + 1)}{\sqrt{x ^ 2 + 1}}[/tex] (make the substitution u = x2 + 1)
    [tex]= \frac{1}{2} \int \frac{(x ^ 2 + 1 - 1) d(x ^ 2 + 1)}{\sqrt{x ^ 2 + 1}} = \frac{1}{2} \int \sqrt{x ^ 2 + 1} d(x ^ 2 + 1) - \frac{1}{2} \int \frac{d(x ^ 2 + 1)}{\sqrt{x ^ 2 + 1}}[/tex].
    Can you go from here? :)
    --------------------
    About arildno's method:
    Integration by Parts is:
    [tex]\int u dv = uv - \int v du[/tex]
    [tex]\int \frac{x ^ 3 dx}{\sqrt{x ^ 2 + 1}}[/tex]
    We then let u = x2, and [tex]dv = \frac{x dx}{\sqrt{x ^ 2 + 1}}[/tex]
    u = x2 => du = 2x dx
    [tex]dv = \frac{x dx}{\sqrt{x ^ 2 + 1}} \Rightarrow v = \sqrt{x ^ 2 + 1}[/tex]
    Substitute all those to our formula, and we are done. Can you get this? :)
     
    Last edited: Jun 29, 2006
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