\(\int \frac{5 \tan x}{\tan x-2} \mathrm{~d} x=x+\mathrm{a} \log |\sin x-2 \cos x|+\mathrm{c}\)
(where \(\mathrm{c}\) is a constant of integration), then the value of \(a\) is

\(\text { Let } \begin{aligned}
\mathrm{I} & =\int \frac{5 \tan x}{\tan x-2} \mathrm{~d} x \\
& =\int \frac{5 \sin x}{\sin x-2 \cos x} \mathrm{~d} x
\end{aligned}\)
Let \(5 \sin x=\mathrm{A}(\sin x-2 \cos x)\)
\(+\mathrm{B} \cdot \frac{\mathrm{d}}{\mathrm{d} x}(\sin x-2 \cos x)\)
\(\begin{aligned}
& \therefore \quad 5 \sin x=\mathrm{A}(\sin x-2 \cos x)+\mathrm{B}(\cos x+2 \sin x) \\
& \therefore \quad \mathrm{A}+2 \mathrm{~B}=5 \text { and }-2 \mathrm{~A}+\mathrm{B}=0
\end{aligned}\)
Solving these equations, we get
\(\begin{aligned}
\mathrm{A} & =1, \mathrm{~B}=2 \\
\therefore \quad \mathrm{I} & =\int 1 \mathrm{~d} x+2 \int \frac{\cos x+2 \sin x}{\sin x-2 \cos x} \mathrm{~d} x \\
& =x+2 \log |\sin x-2 \cos x|+\mathrm{c}
\end{aligned}\)
But \(\int \frac{5 \tan x}{\tan x-2} \mathrm{~d} x=x+\operatorname{alog}|\sin x-2 \cos x|+\mathrm{c}\)
Comparing, we get \(\mathrm{a}=2\)