1. ## Integral help...

If $f(x) = \int_{0}^{g(x)}\frac{1}{\sqrt{1+t^3}}dt$, where $g(x) = \int_{0}^{cos(x)} (1 + sin(t^2))dt$, find $f ' (\frac{\pi}{2})$.

Help would be much appreciated.
I know this is mostly a problem when I need to know how integrals work, but I'm having a hard time reasoning it out. Could someone explain how it's done and give me an answer to check with?

2. Originally Posted by ebonyscythe
If $f(x) = \int_{0}^{g(x)}\frac{1}{\sqrt{1+t^3}}dt$, where $g(x) = \int_{0}^{cos(x)} (1 + sin(t^2))dt$, find $f ' (\frac{\pi}{2})$.

Help would be much appreciated.
I know this is mostly a problem when I need to know how integrals work, but I'm having a hard time reasoning it out. Could someone explain how it's done and give me an answer to check with?
use the second fundamental theorem of calculus,,

3. Thanks for the hint... I'm going to try and work through this one now that I have a chance.

I'm still hoping for an answer to compare to eventually, or some help reasoning through the problem if someone has the time...

4. Okay, well I tried the problem and I'm still having a hard time reasoning anything out. I've got the 2nd fund theorem of calc in front of me telling me that:

If function S is continuous on [a,b], then the function D is defined by
$D(x)=\int_{a}^{x}S(t)dt$ where $a\le x\le b$
and is continuous on [a,b], differentiable on (a,b) and
$D ' (x)=S(x)$

So, would that mean, in my problem above, that $f'(x)=\frac{1}{\sqrt{1+g(x)^3}}*g'(x)$?

5. Originally Posted by ebonyscythe
Okay, well I tried the problem and I'm still having a hard time reasoning anything out. I've got the 2nd fund theorem of calc in front of me telling me that:

If function S is continuous on [a,b], then the function D is defined by
$D(x)=\int_{a}^{x}S(t)dt$ where $a\le x\le b$
and is continuous on [a,b], differentiable on (a,b) and
$D ' (x)=S(x)$

So, would that mean, in my problem above, that $f'(x)=\frac{1}{\sqrt{1+g(x)^3}}*g'(x)$?
yes! and what is $g'(x)$?

6. Would $g'(x) = -sin(x)[1+sin(cos^2x)]?$

If so, how do I put this thing together? I can't imagine taking the third root of the integral g(x)...

Would I just push the $\pi/2$ through? and if so, is $f'(\pi/2)=-1 ?$

7. Originally Posted by ebonyscythe
Would $g'(x) = -sin(x)[1+sin(cos^2x)]?$

If so, how do I put this thing together? I can't imagine taking the third root of the integral g(x)...

Would I just push the $\pi/2$ through? and if so, is $f'(\pi/2)=-1 ?$
yes, you just have to plug in $\frac{\pi}{2}$ on every $x$ you see.., for the case of $g(x)$, you should notice the the upper limit of the integral would become real and therefore, $g(x)$ can "easily" be integrated.. thus, its integral replaces the $g(x)$ on your $f'(x)$ and then, you are done..

8. Thanks for all of your help!