
#1
Aug1211, 01:10 PM

P: 74

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 [tan^8(sin^2(x))]/sin^18(x) my book says to use l'hospital's rule, so i continued with [8tan^7(sin^2x)*sec^2(sin^2(X))*2sinxcosx] but my book says i should just have [tan^8(sin^2x)*2sinxcosx]/18sin^17(x)cos(x) why did they only partially differentiate tan^8(sin^2(x))? 



#2
Aug1211, 01:40 PM

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P: 21,012

[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. 



#3
Aug1211, 01:46 PM

P: 74

[tex]\lim_{x \to0} 1/{sin^18(x)} \int_0^{sin^2(x)} tan^8(t)~dt[/tex] 



#4
Aug1211, 01:49 PM

P: 74

l'hospital's rule with trig functionsbut it's 1/[sin^18(x)] 



#5
Aug1211, 01:57 PM

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P: 21,012

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



#6
Aug1211, 02:04 PM

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P: 21,012

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] 



#7
Aug1211, 02:06 PM

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put 18 inside {} → ^{18}




#8
Aug1211, 02:35 PM

P: 74




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