The function is $$F(x)=\int_{1}^{4x}\cot(t)\,dt. $$
I begin by writing in terms of $\sin$ and $\cos$ to make substitution easier. $$F(x)=\int_{1}^{4x} \frac{\cos(t)}{\sin(t)}\,dt. $$ From here I picked my $u$ substitution \begin{align*} u&=\sin(u)\\ du&=\cos(t)\,dt, \end{align*} which yielded $$\int_{1}^{4x} \frac{1}{u}\,du,$$ and thus I get $$\ln|\sin(t)|. $$ My answer should be $4\cot(4x)$ and I can't figure out how to get a $\cos$ from this or if I messed up earlier on.
The idea is to use the first part of the FTC.
Note that $F=f\circ g$, where $$f(x)=\int_1^x \cot(t)\,dt \qquad \text{and}\qquad g(x)=4x.$$ Then $$f'(x)=\cot(x) \qquad \text{and}\qquad g'(x)=4.$$ Therefore, by the chain rule, the derivative of $F(x)$ is $f'(g(x))\,g'(x)$. That is, $$F'(x)=\cot(4x)\cdot4.$$