# Let $F(x)=cos\int_{0}^{e^{x}}cos^{2}(\int_{0}^{u}(cos^{3}t)dt)du$ Find $F'(x)$

Let $F(x)=cos\int_{0}^{e^{x}}cos^{2}(\int_{0}^{u}(cos^{3}t)dt)du$, Find $F'(x)$

I have thought to solve the exercise using the fundamental theorem of the calculation and the rule of the chain

I know this:

$H(x)=\int_{0}^{u}cos^{3}t\Rightarrow H'(x)=cos^{3}u$

$G(x)=\int_{0}^{e^{x}}cos^{2}(\int_{0}^{u}cos^{3}t)\Rightarrow > G'(x)=(?)$

How can I derive integrals within integrals?

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And how i can solve $F(x)$? With the same procedure?

We write $$F(x)=\cos(g(x))$$ with $$g(x)=\int_0^{e^x}\cos^2\left(\int_0^u\cos^3(t)dt\right)du.$$
By the chain rule, $$F'(x)=-\sin(g(x))g'(x)$$
We can write $$g(x)=\int_0^{e^x}\cos^2(H(u))du$$ with $H(u)=\int_0^u\cos^3(t)dt$.
Now, suppose $G(x)$ is such that $G'(x)=\cos^2(H(x))$. Then we can write $g(x)=G(e^x)-G(0)$. Thus, $$g'(x)=G'(e^x)e^x=\cos^2(H(e^x))e^x.$$
Putting everything together, we have $$F'(x)=-\sin\left(\int_0^{e^x}\cos^2\left(\int_0^u\cos^3(t)dt\right)du\right)\cos^2\left(\int_0^{e^x}\cos^3(t)dt\right)e^x$$
Let $$q(u)=cos^{2}(\int_{0}^{u}(cos^{3}t)dt)$$ Then $$F(x)=cos(\int_{0}^{e^{x}}q(u)du)$$ so $$F'(x)=-sin(\int_{0}^{e^{x}}q(u)du)e^xq(e^x)$$ which you can expand to give the answer. I hope that this isn't a homework problem; if so, the person who assigned it just wanted to watch students make a mistake when writing out the answer.