if the integral $\int \frac{5\tan \mathrm{x}}{\tan \mathrm{x}-2}\mathrm{dx}=\mathrm{x}+\mathrm{alog}$ $|\sin x-2\cos x|+k\text{, then }a\text{ is equal to }$

Given integral $\int \frac{5\tan \mathrm{x}}{\tan \mathrm{x}-2}\mathrm{dx}$

To find The value of $\mathrm{a}\text{ ', if }\int \frac{5\tan \mathrm{x}}{\tan \mathrm{x}-2}\mathrm{dx}$ 

$=\mathrm{x}+\mathrm{alog}|\sin \mathrm{x}-2\cos \mathrm{x}|+\mathrm{k} ---\left(\mathrm{i}\right)$

Now, let us assume that $\mathrm{I}=\int \frac{5\tan \mathrm{x}}{\tan \mathrm{x}-2}\mathrm{dx}$

On multiplying by cos x in numerator and
denominator, we get 

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

This special integration requires special substitution
of type

${\mathrm{N}}^{\mathrm{r}}=\mathrm{A}\left({\mathrm{D}}^{\mathrm{r}}\right)+\mathrm{B}\left(\frac{{\mathrm{dD}}^{\mathrm{r}}}{\mathrm{dx}}\right),\mathrm{A}\text{ and }\mathrm{B}$ are constant.

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

$\Rightarrow 0\cos \mathrm{x}+5\sin \mathrm{x}=(\mathrm{A}+2\mathrm{B})\sin \mathrm{x}+(\mathrm{B}-2\mathrm{A})\cos \mathrm{x}$

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

On solving the above two equations in A and B, we get

A=1 and  B=2 

$\begin{array}{l}\Rightarrow 5\sin \mathrm{x}=(\sin \mathrm{x}-2\cos \mathrm{x})+2(\cos \mathrm{x}+2\sin \mathrm{x})\\ \Rightarrow \mathrm{I}=\int \frac{5\sin \mathrm{x}}{\sin \mathrm{x}-2\cos \mathrm{x}}\mathrm{dx}\\ =\int \frac{(\sin \mathrm{x}-2\cos \mathrm{x})+2(\cos \mathrm{x}+2\sin \mathrm{x})}{(\sin \mathrm{x}-2\cos \mathrm{x})}\mathrm{dx}\\ =\int \frac{\sin \mathrm{x}-2\cos \mathrm{x}}{\sin \mathrm{x}-2\cos \mathrm{x}}\mathrm{dx}+2\int \frac{(\cos \mathrm{x}+2\sin \mathrm{x})}{(\sin \mathrm{x}-2\cos \mathrm{x})}\mathrm{dx}\\ =\int 1\mathrm{dx}+2\int \frac{\mathrm{d}(\sin \mathrm{x}-2\cos \mathrm{x})}{(\sin \mathrm{x}-2\cos \mathrm{x})}\end{array}$

$=\mathrm{x}+2\log \left|\right(\sin \mathrm{x}-2\cos \mathrm{x}\left)\right|+\mathrm{k} ----\left(\mathrm{ii}\right)$

where, k is the constant of integration.
Now, by comparing the value of I in Eqs. (i) and (ii),

we get .