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

Solution :
$$\eqalign{ & \int {\frac{{5\,\tan \,x}}{{\tan \,x - 2}}dx} = \int {\frac{{5\frac{{\sin \,x}}{{\cos \,x}}}}{{\frac{{\sin \,x}}{{\cos \,x}} - 2}}dx} \cr & = \int {\left( {\frac{{5\sin \,x}}{{\cos \,x}} \times \frac{{\cos \,x}}{{\sin \,x - 2\,\cos \,x}}} \right)dx} \cr & = \int {\frac{{5\sin \,x\,dx}}{{\sin \,x - 2\cos \,x}}} \cr & = \int {\left( {\frac{{4\sin \,x + \sin \,x + 2\cos \,x - 2\cos \,x}}{{\sin \,x - 2\cos \,x}}} \right)} dx \cr & = \int {\left( {\frac{{\left( {\sin \,x - 2\cos \,x} \right) + \left( {4sin\,x + 2\cos \,x} \right)}}{{\sin \,x - 2\cos \,x}}} \right)} dx \cr & = \int {\left( {\frac{{\left( {\sin \,x - 2\cos \,x} \right) + 2\left( {\cos \,x + 2\sin \,x} \right)}}{{\sin \,x - 2\cos \,x}}} \right)} dx \cr & = \int {\frac{{\sin \,x - 2\,\cos \,x}}{{\sin \,x - 2\cos \,x}}} dx + 2\int {\left( {\frac{{\cos \,x + 2\sin \,x}}{{\sin \,x - 2\cos \,x}}} \right)dx} \cr & = \int {dx + 2} \int {\frac{{\cos \,x + 2\sin \,x}}{{\sin \,x - 2\cos \,x}}dx} \cr & = {I_1} + {I_2} \cr & {\text{Where}}\,\,{I_1} = \int {dx} {\text{ and}} \cr & {I_2} = 2\int {\frac{{\cos \,x + 2\sin \,x}}{{\sin \,x - 2\cos \,x}}} dx \cr & {\text{Put }}\sin \,x - 2\cos \,x = t \cr & \Rightarrow \left( {\cos \,x + 2\sin \,x} \right)dx = dt \cr & \therefore {I_2} = 2\int {\frac{{dt}}{t} = 2\ln \,t + C = 2\,\ln \,} \left( {\sin \,x - 2\cos \,x} \right) + C \cr & {\text{Hence,}}\,\,{I_1} + {I_2} = \int {dx + 2\ln \left( {\sin \,x - 2\cos \,x} \right) + c} \cr & = x + 2\ln \left|( {\sin \,x - 2\cos \,x} \right)| + k\,\,\, \Rightarrow a = 2 \cr} $$