# L'hospital's rule with trig functions

1. Aug 12, 2011

### Shannabel

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. Aug 12, 2011

### Staff: Mentor

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?

3. Aug 12, 2011

### Shannabel

no, it should be
$$\lim_{x \to0} 1/{sin^18(x)} \int_0^{sin^2(x)} tan^8(t)~dt$$

Last edited: Aug 12, 2011
4. Aug 12, 2011

### Shannabel

well i can't make it look right..
but it's 1/[sin^18(x)]

5. Aug 12, 2011

### 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.

6. Aug 12, 2011

### 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.
$$\frac{d}{dx} \int_a^x f(t)~dt~= f(x)$$

If the limit of integration contains a function of x, then you need to use the chain rule.
$$\frac{d}{dx} \int_a^{g(x)} f(t)~dt~= \frac{d}{du} \int_a^{u} f(t)~dt * \frac{du}{dx}~= f(u) * du/dx$$

7. Aug 12, 2011

### SammyS

Staff Emeritus
put 18 inside {} → ^{18}
Is this the same as: $$\lim_{x \to0} \frac{ \int_0^{sin^2(x)} tan^8(t)~dt}{sin^{18}(x)} \ ?$$

8. Aug 12, 2011

### Shannabel

that makes much more sense, thanks!