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L'hospital's rule with trig functions

  1. Aug 12, 2011 #1
    1. The problem statement, all variables and given/known data
    evaluate lim(x->0) (tan^8(t))dt(between 0 and sin^2x)

    2. Relevant equations

    3. The attempt at a solution

    my book says to use l'hospital's rule, so i continued with

    but my book says i should just have

    why did they only partially differentiate tan^8(sin^2(x))?
  2. jcsd
  3. Aug 12, 2011 #2


    Staff: Mentor

    Is this supposed to be a definite integral? If so, is this the problem?
    [tex]\lim_{x \to 0} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]

    You can see the LaTeX I used by clicking on the integral.
    How did you get this?
  4. Aug 12, 2011 #3
    no, it should be
    [tex]\lim_{x \to0} 1/{sin^18(x)} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]
    Last edited: Aug 12, 2011
  5. Aug 12, 2011 #4
    well i can't make it look right..
    but it's 1/[sin^18(x)]
  6. Aug 12, 2011 #5


    Staff: Mentor

    Fixed it.
    Tip: In LaTeX, if an exponent has more than one character (and that includes pos. or negative sign) put braces {} around it.
  7. Aug 12, 2011 #6


    Staff: Mentor

    OK, that makes more sense. L'Hopital's is applicable here.

    In answer to your questions, they didn't "partially differentiate" the expression in the integrand - they used the chain rule.

    You know (I think) this part of the Fund. Thm. of Calculus involving derivatives and definite integrals.
    [tex]\frac{d}{dx} \int_a^x f(t)~dt~= f(x)[/tex]

    If the limit of integration contains a function of x, then you need to use the chain rule.
    [tex]\frac{d}{dx} \int_a^{g(x)} f(t)~dt~= \frac{d}{du} \int_a^{u} f(t)~dt * \frac{du}{dx}~= f(u) * du/dx[/tex]
  8. Aug 12, 2011 #7


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    put 18 inside {} → ^{18}
    Is this the same as: [tex]\lim_{x \to0} \frac{ \int_0^{sin^2(x)} tan^8(t)~dt}{sin^{18}(x)} \ ?[/tex]
  9. Aug 12, 2011 #8
    that makes much more sense, thanks!
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