If the integral $\int \frac{5\tan x}{\tan x-2}dx=x+a\ln |\sin x-2\cos x|+k$, then $a$ is equal to:

$\int \frac{5\tan x}{\tan x-2}dx=\int \frac{5\frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x}-2}dx=\int \frac{5\sin x}{\sin x-2\cos x}dx$ Let the numerator be expressed as:$5\sin x=A(\sin x-2\cos x)+B\frac{d}{dx}(\sin x-2\cos x)$$5\sin x=A(\sin x-2\cos x)+B(\cos x+2\sin x)$ Equating coefficients of $\sin x$: $A+2B=5$ Equating coefficients of $\cos x$: $-2A+B=0⟹B=2A$ Substitute $B=2A$ into the first equation:$A+2(2A)=5⟹5A=5⟹A=1$ Then $B=2$.The integral becomes:$\int \left(1+\frac{2(\cos x+2\sin x)}{\sin x-2\cos x}\right)dx$$=\int 1dx+2\int \frac{\cos x+2\sin x}{\sin x-2\cos x}dx$$=x+2\ln |\sin x-2\cos x|+k$ Comparing with $x+a\ln |\sin x-2\cos x|+k$, we get $a=2$.