We are asked to evaluate two integrals using the given substitutions. The integrals are: (a) $\int \frac{\cos(\ln t)}{t} dt$, with $u = \ln t$. (b) $\int_e^{e^4} \frac{dx}{x \sqrt{\ln x}}$, with $u = \ln x$.

1. Problem Description

We are asked to evaluate two integrals using the given substitutions. The integrals are:

(a) $\int \frac{\cos(\ln t)}{t} dt$, with $u = \ln t$.

(b) $\int_e^{e^4} \frac{dx}{x \sqrt{\ln x}}$, with $u = \ln x$.

2. Solution Steps

(a) We are given $\int \frac{\cos(\ln t)}{t} dt$ and $u = \ln t$. Then $\frac{du}{dt} = \frac{1}{t}$, so $du = \frac{1}{t} dt$. Substituting into the integral, we have

$\int \frac{\cos(\ln t)}{t} dt = \int \cos(u) du = \sin(u) + C = \sin(\ln t) + C$.

(b) We are given $\int_e^{e^4} \frac{dx}{x \sqrt{\ln x}}$ and $u = \ln x$. Then $\frac{du}{dx} = \frac{1}{x}$, so $du = \frac{1}{x} dx$. The integral becomes $\int \frac{du}{\sqrt{u}}$.

When $x=e$, $u = \ln e = 1$. When $x=e^4$, $u = \ln(e^4) = 4$. So the new limits of integration are from 1 to

4. $\int_1^4 \frac{1}{\sqrt{u}} du = \int_1^4 u^{-1/2} du = [2u^{1/2}]_1^4 = 2\sqrt{4} - 2\sqrt{1} = 2(2) - 2(1) = 4-2 = 2$.

3. Final Answer

(a) $\sin(\ln t) + C$

(b) $2$