Step 1: Understand the goal.
We must find $a$ given that \[ \int\frac{5\tan x}{\tan x - 2}\,dx = x + a\log|\sin x - 2\cos x| + c. \] The plan is to rewrite the integrand so part of it is a constant and part is a derivative over a function.
Step 2: Write tangent as sine over cosine.
\[ \frac{5\tan x}{\tan x - 2} = \frac{5\sin x}{\sin x - 2\cos x}. \] We multiplied top and bottom by $\cos x$ to remove the tangents.
Step 3: Match the numerator to the denominator and its derivative.
The denominator is $D = \sin x - 2\cos x$. Its derivative is $D' = \cos x + 2\sin x$. We try to write $5\sin x = A\,D + B\,D'$ for constants $A$ and $B$.
Step 4: Solve for A and B.
So $5\sin x = A(\sin x - 2\cos x) + B(\cos x + 2\sin x)$. Matching $\sin x$: $A + 2B = 5$. Matching $\cos x$: $-2A + B = 0$, giving $B = 2A$. Then $A + 4A = 5$, so $A = 1$ and $B = 2$.
Step 5: Split the integral.
\[ \int\frac{A\,D + B\,D'}{D}\,dx = \int A\,dx + \int B\,\frac{D'}{D}\,dx = A\,x + B\log|D|. \]
Step 6: Read off $a$.
This gives $x + 2\log|\sin x - 2\cos x| + c$. Comparing with the given form, $a = 2$. \[ \boxed{a = 2} \]