Find the derivative of $F(x)=\int_{\pi}^{\ln x} \cos e^t dt$ Am I supposed to change the limits of integration?
$$F(x)=\int_{\pi}^{\ln x} \cos e^t dt = \int_{e^\pi}^x \frac{\cos u}{u} du $$
Help!
 A: Since $\frac{d}{dx}\int_a^xg(u)du=g(x)$, the answer is $\frac{\cos x}{x}$.
A: The leibniz integral rule states that:
$$\frac{d}{dx}\int_{a(x)}^{b(x)}f(x,t)dt=f(x,b(x))\,b'(x)-f(x,a(x))\,a(x)+\int_{a(x)}^{b(x)}\partial_xf(x,t)dt$$
so we can say that if:
$$F(x)=\int_\pi^{\ln(x)}\cos(e^t)dt$$
then:
$$F'(x)=\cos(e^{\ln(x)})\frac 1x=\frac{\cos(x)}{x}$$
A: This question is really an application of the fundamental theorem of calculus and the chain rule. To make it more transparent, define the following "auxillary" functions:
\begin{equation}
G(x) = \int_{\pi}^x \cos(e^t) \, dt
\end{equation}
and
\begin{equation}
\psi(x) = \ln(x)
\end{equation}
Then, the function $F$ you're interested in can be written as a composition $F(x) = (G \circ \psi)(x)$. The chain rule first tells you that
\begin{equation}
F'(x) = G'(\psi(x)) \cdot \psi'(x)
\end{equation}
$\psi'(x) = \dfrac{1}{x}$ should be known to you. The fundamental theorem of calculus says that $G'(x) = \cos(e^x)$. Putting this all together (and being careful about where you evaluate derivatives) you get:
\begin{align}
F'(x) &= G'(\psi(x)) \cdot \psi'(x) \\
&= \cos(e^{\psi(x)}) \cdot \dfrac{1}{x} \\
&:= \cos(e^{\ln(x)}) \cdot \dfrac{1}{x} \\
&= \cos(x) \cdot \dfrac{1}{x}
\end{align}

As an additional remark, whenever you have a function defined by an integral, such as 
\begin{equation}
F(x) = \int_a^{\psi(x)} g(t) \, dt
\end{equation}
where $a$ is a constant, and $g, \psi$ are functions (with enough regularity, such as $g$ continuous and $\psi$ differentiable), to compute $F'(x)$, you should resort to the method above; DO NOT try to first calculate the integral, then take the derivative, because sometimes it can be difficult, and even impossible. If you systematically apply the chain rule and Fundamental theorem of calculus, you should find that
\begin{equation}
F'(x) = g(\psi(x)) \cdot \psi'(x)
\end{equation}
Also, yes, changing the limits is fine in this example, but this was only possible because you knew the explicit relationship that $\exp$ and $\ln$ are inverse functions. More generally, it may difficult to find the inverse explicitly, and in such a case, using the chain rule method is far quicker
A: I'd argue that changing the limits of integration is fine, but not needed. I usually employ the following approach: the integrand has some antiderivative, which I'll call $G(t)$. Then evaluating the definite integral gives:
$$G(\ln x)-G(\pi).$$
Differentiating this with respect to $x$ gives:
$$\dfrac{G'(\ln x)}{x},$$
since $G(\pi)$ is a constant, and we know $G'(x)=\cos(e^x)$, so our final answer is:
$$\dfrac{\cos(e^{\ln x})}{x}=\dfrac{\cos x}{x}$$
