# Definite Integration with Upper bound as another integral

1. Aug 22, 2012

### marathon

i have a similar one.

f(x) = $\int$$\frac{dt}{\sqrt{1+t^3}}$ on (0, g(x))

g(x) = $\int$(1+sin(t^2))dt on (0, cos(x))

that is, these are definite integrals on the interval from zero up to the given function.

the question is to solve f'(pi/2). the correct answer is -1 but i don't understand how to get it. i'm aware that integrating sin(t^2) leads to a fresnel, but i'm pretty sure that's not what we're supposed to be doing here since it's a challenge problem from a calc 2 text on the chapter with FTC. what i got was that g(pi/2) should be 0 since cos(pi/2)=0 and any integral on (a,a)=0. then we'd have f'(pi/2) = $\frac{1}{\sqrt{1+g(\frac{\pi}{2})^3}}$=$\frac{1}{\sqrt{1+0^3}}$ = 1, but obviously this is incorrect. halp?

2. Aug 22, 2012

### Dick

Re: Definite Integration with Upper bound as another integral.

Look at the Leibniz integral rule, http://en.wikipedia.org/wiki/Leibniz_integral_rule in the section under 'Variable limits'. You need to multiply by a derivative of the upper limit with respect to x.