# l'hospital's rule with trig functions

by Shannabel
Tags: functions, lhospital, rule, trig
 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))?
Mentor
P: 20,436
 Quote by Shannabel 1. The problem statement, all variables and given/known data evaluate lim(x->0) (tan^8(t))dt(between 0 and sin^2x)
Is this supposed to be a definite integral? If so, is this the problem?
$$\lim_{x \to 0} \int_0^{sin^2(x)} tan^8(t)~dt$$

You can see the LaTeX I used by clicking on the integral.
 Quote by Shannabel 2. Relevant equations 3. The attempt at a solution [tan^8(sin^2(x))]/sin^18(x)
How did you get this?
 Quote by Shannabel 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))?
P: 74
 Quote by Mark44 Is this supposed to be a definite integral? If so, is this the problem? $$\lim_{x \to 0} \int_0^{sin^2(x)} tan^8(t)~dt$$ You can see the LaTeX I used by clicking on the integral. How did you get this?
no, it should be
$$\lim_{x \to0} 1/{sin^18(x)} \int_0^{sin^2(x)} tan^8(t)~dt$$

P: 74

## l'hospital's rule with trig functions

 Quote by Shannabel no, it should be $$\lim_{x \to0} 1/{sin^18(x)} \int_0^{sin^2(x)} tan^8(t)~dt$$
well i can't make it look right..
but it's 1/[sin^18(x)]
Mentor
P: 20,436
Fixed it.
Tip: In LaTeX, if an exponent has more than one character (and that includes pos. or negative sign) put braces {} around it.
 Quote by Shannabel no, it should be $$\lim_{x \to 0} \frac{1}{sin^{18}(x)} \int_0^{sin^2(x)} tan^8(t)~dt$$
 Mentor P: 20,436 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. $$\frac{d}{dx} \int_a^x f(t)~dt~= f(x)$$ If the limit of integration contains a function of x, then you need to use the chain rule. $$\frac{d}{dx} \int_a^{g(x)} f(t)~dt~= \frac{d}{du} \int_a^{u} f(t)~dt * \frac{du}{dx}~= f(u) * du/dx$$
Emeritus
 Quote by Shannabel no, it should be $$\lim_{x \to0} 1/{sin^{18}(x)} \int_0^{sin^2(x)} tan^8(t)~dt$$
Is this the same as: $$\lim_{x \to0} \frac{ \int_0^{sin^2(x)} tan^8(t)~dt}{sin^{18}(x)} \ ?$$
 Quote by Mark44 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. $$\frac{d}{dx} \int_a^x f(t)~dt~= f(x)$$ If the limit of integration contains a function of x, then you need to use the chain rule. $$\frac{d}{dx} \int_a^{g(x)} f(t)~dt~= \frac{d}{du} \int_a^{u} f(t)~dt * \frac{du}{dx}~= f(u) * du/dx$$