Evaluating $\;\int x\cos(5x-1)\,\mathrm{d}x$ Integration by parts:
1) let $u = x$, $du = dx$, $v = \frac{1}{5}  \sin (5x- 1)$, $dv = \cos (5x -1) dx$
$udv = x \frac{1}{5} \sin x ( 5x- 1) - \frac{1}{5} \int \sin (5x-1) dx$.
2) let $u = \sin (5x -1), du = \frac{1}{5} \cos(5x-1) dx , v = x, dv = dx$
Is my $du$ correct in part 2?

The expression should be $$I = \int u\,dv = uv -\int v\,du $$
which is what I suspect you meant, since you correctly obtained
$$I = \frac 15 x \sin(5x - 1) - \frac 15\int \sin(5x - 1)\,dx $$
From there, using substitution (see note below), we get
$$ I = \frac 15 x \sin(5x - 1) - \frac 1{25}( - \cos(5x - 1))+ C$$
$$= \frac 15 x \sin(5x - 1) + \frac 1{25}\cos(5x - 1)+ C$$

Note: Integrating $\frac 15 \int \sin(5x-1)\,dx$ doesn't require using integration by parts. Rather, you can use substitution putting $u = 5x-1\implies du = 5\,dx \iff \frac 15\,du = dx$ gives you 
$$\frac 15 \int \sin u \left(\frac 15 \,du \right) = \frac 1{25} \int \sin u \,du$$ $$ = \frac 1{25}(-\cos u) + C = -\frac 1{25}\cos(5x-1) + C$$
