1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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...


    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Don't follow??
    It is perfectly obvious!
    We get [itex]v=\sqrt{x^{2}+1}[/itex], thus:
    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.

    [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

    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


    User Avatar
    Homework Helper

    This line is incorrect.
    You should note that:
    [tex]\int x ^ {\alpha} dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C, \ \alpha \neq -1[/tex]
    [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]
    [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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook