FTC integrate multiple functions.

[tropic9393]

Homework Statement

If F(x)=∫from 0 to g(x) of 1/(√(1+t^3)) dt and g(x)= ∫from 0 to cos(x) of 1+sin(t^2) dt, then find f'(pi/2)

Homework Equations

The Attempt at a Solution

Tried FTC parts 1 and 2

 

[SammyS]

Homework Statement

If F(x)=∫from 0 to g(x) of 1/(√(1+t^3)) dt and g(x)= ∫from 0 to cos(x) of 1+sin(t^2) dt, then find f'(pi/2)

Homework Equations

The Attempt at a Solution

Tried FTC parts 1 and 2

Hello tropic9393. Welcome to PF !

Please, show us explicitly what you have tried so that we may help you.

 

[tropic9393]

I know I need to solve for g(x). I did the integral with respect to t of and got realized that sin(t^2) does not have an elementary integral, and I haven't learned how to do solve for that yet.
Then i tried applying FTC directly and but i thought that only applied to the derivative of an integral.
I have a major conceptual block, so i don't really have a lot of work to show

 

[SammyS]

I know I need to solve for g(x). I did the integral with respect to t of and got realized that sin(t^2) does not have an elementary integral, and I haven't learned how to do solve for that yet.
Then i tried applying FTC directly and but i thought that only applied to the derivative of an integral.
I have a major conceptual block, so i don't really have a lot of work to show

You are correct regarding the anti-derivative of sin(t²).

Neither does $\displaystyle 1/\sqrt{1+t^3}\$ have an anti-derivative which can be expressed in terms of elementary functions.

Approach the solution in two parts, similar to the way the problem is set-up.

According to the Fundamental Theorem of Calculus, what is f '(x), if $\displaystyle f(x)=\int_{0}^{g(x)}\frac{1}{\sqrt{1+t^3}}\,dt\ ?$ (Without using the explicit form of g(x). )

 

[tropic9393]

f'(x)=1/(√(1+g(x)^3)

Then do solve for f'(pi/2) by plugging g(pi/2) in for g(x)?

 

[SammyS]

f'(x)=1/(√(1+g(x)^3)

...

Not quite right. You forgot to use the chain rule.
Suppose that the anti-derivative of $1/\sqrt{1+t^3}$ is $H(t)$.

That says that $\displaystyle \int_{0}^{g(x)}\frac{1}{\sqrt{1+t^3}}\,dt=H(g(x))-H(0)\ .$

Use the chain rule to find the derivative, $\displaystyle \frac{d}{dx}H(g(x))\ .$