If the integral $\int \frac{5\mathrm{tanx}}{\mathrm{tanx}-2}\mathrm{dx}=\mathrm{x}+\mathrm{a}\text{\hspace{0.17em}l}n|\mathrm{sinx}-2\mathrm{cosx}|+\mathrm{k},$ then a is equal to 

$I=\int \frac{5\mathrm{tanx}}{\mathrm{tanx}-2}dx$

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

$$\begin{gathered} =\int \frac{0\cos x+5\sin x}{-2\cos x+\sin x}dx \\ Formula \\ \int \frac{a \cos x+b \sin x}{c \cos x+d \sin x}dx=Ax+B \log \left|C \cos x+d \sin x\right| \\ Where A=\frac{ac+bd}{c^2+d^2}, B=\frac{ad-bc}{c^2+d^2} \\ here A=\frac{0\left(-2\right)+5\left(1\right)}{4+1}=1, B=\frac{0+10}{4+1}=2 \\ \therefore I=1.x+2\log \left|-2\cos x+Sinx\right|+k \\ a=2 \end{gathered}$$