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Definite Integration with Upper bound as another integral

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i have a similar one.

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

g(x) = [itex]\int[/itex](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) = [itex]\frac{1}{\sqrt{1+g(\frac{\pi}{2})^3}}[/itex]=[itex]\frac{1}{\sqrt{1+0^3}}[/itex] = 1, but obviously this is incorrect. halp?
 

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Dick
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