If the integral $\int \frac{5\tan x}{\tan x-2}dx=x+$$a\log |\sin x-2\cos x|+\mid K$ then a is equal to

$I=\int \frac{5\tan x}{\tan x-2}dx=\int \frac{5\sin x}{\sin x-2\cos x}dx$

Let $5\sin x=A(\sin x-2\cos x)+B(\cos x+2\sin x)$

$\text{ Equating coefficients of }\sin x\text{ and }\cos x\text{, we get }$

$5=A+2B$ and $0=-2A+B$

$\begin{array}{l}\Rightarrow A=1,B=2\\ \therefore I=\int \frac{(\sin x-2\cos x)+2(\cos x+2\sin x)}{\sin x-2\cos x}dx\\ =x+2\log |\sin x-2\cos x|+K\\ \therefore a=2\end{array}$