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

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

$\text{Put }5\sin x=A\frac{\mathrm{d}}{\mathrm{d}x}(\sin x-2\cos x)+B(\sin x-2\cos x)$

                $=A(\cos x+2\sin x)+b(\sin x-2\cos x)$

                $=A(\cos x+2\sin x)+b(\sin x-2\cos x)$

$\text{ Comparing like coefficients on both sides }$

             $5=2A+B \text{ and }0=A-2B$

          $\therefore A=2,B=1$

$Now,$

$\int \frac{5\sin x}{\sin x-2\cos x} \mathrm{d}x=2\int \frac{\cos x+2\sin x}{(\sin x-2\cos x)}dx+\int \frac{\sin x-2\cos x}{\sin x-2\cos x} \mathrm{d}x$

                                $=2\log (\sin x-2\cos x)+x+K$

                                $=x+a\log (\sin x-2\cos x)+K\text{ (given) }$

$\Rightarrow a=2$