If the integral, $\int {\dfrac{{5\tan x}}{{\tan x - 2}}dx = x + a\ln \left| {\sin x - 2\cos x} \right| + k} $, then a is equal to:
$\left( a \right)$ 1
$\left( b \right)$ -2
$\left( c \right)$ -1
$\left( d \right)$ 2

Hint: In this particular question use the concept that tan x = (sin x/cos x) so according to this property simplify the integral then write the numerator of the simplified integral as 5 sin x = {(sin x – 2 cos x) + 2 (cos x + 2sin x)}, then separate the integral so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given integral
$\int {\dfrac{{5\tan x}}{{\tan x - 2}}dx = x + a\ln \left| {\sin x - 2\cos x} \right| + k} $
Consider the LHS of the above equation we have,
\[ \Rightarrow \int {\dfrac{{5\tan x}}{{\tan x - 2}}dx} \]
Let, \[I = \int {\dfrac{{5\tan x}}{{\tan x - 2}}dx} \]
Now as we know that tan x = (sin x/cos x) so use this property in the above equation we have,
Let, \[I = \int {\dfrac{{5\left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{\dfrac{{\sin x}}{{\cos x}} - 2}}dx} \]
Now simplify it we have,
\[ \Rightarrow I = \int {\dfrac{{5\sin x}}{{\sin x - 2\cos x}}dx} \]
Now 5 sin x is written as, 5 sin x = {(sin x – 2 cos x) + 2 (cos x + 2sin x)} so apply this in the above integral we have,
\[ \Rightarrow I = \int {\dfrac{{\left\{ {\left( {\sin x - 2\cos x} \right) + 2\left( {\cos x + 2\sin x} \right)} \right\}}}{{\sin x - 2\cos x}}dx} \]
Now separate the integral we have,
\[ \Rightarrow I = \int {\dfrac{{\sin x - 2\cos x}}{{\sin x - 2\cos x}}dx} + 2\int {\dfrac{{\cos x + 2\sin x}}{{\sin x - 2\cos x}}dx} \]
Now simplify it we have,
\[ \Rightarrow I = \int {1dx} + 2\int {\dfrac{{\cos x + 2\sin x}}{{\sin x - 2\cos x}}dx} \].................. (1)
Now let, $\sin x - 2\cos x = t$................ (2)
Now differentiate equation (2) w.r.t x we have,
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sin x - 2\cos x} \right) = \dfrac{{dt}}{{dx}}$
Now as we know that $\dfrac{d}{{dx}}\sin x = \cos x,\dfrac{d}{{dx}}\cos x = - \sin x$ so we have,
$ \Rightarrow \left( {\cos x + 2\sin x} \right) = \dfrac{{dt}}{{dx}}$
$ \Rightarrow \left( {\cos x + 2\sin x} \right)dx = dt$.................... (3)
Now substitute the values from equation (2) and (3) in equation (1) we have,
\[ \Rightarrow I = \int {1dx} + 2\int {\dfrac{1}{t}dt} \]
Now as we know that $\int {\dfrac{1}{x}dx} = \ln \left| x \right| + k,\int {1dx} = x + k$, where k is some arbitrary integration constant so we have,
\[ \Rightarrow I = x + 2\ln \left| t \right| + k\]
Now substitute the value of k in the above equation we have,
\[ \Rightarrow I = x + 2\ln \left| {\sin x - 2\cos x} \right| + k\]
$ \Rightarrow \int {\dfrac{{5\tan x}}{{\tan x - 2}}dx = x + 2\ln \left| {\sin x - 2\cos x} \right| + k} $
Now on comparing with, $\int {\dfrac{{5\tan x}}{{\tan x - 2}}dx = x + a\ln \left| {\sin x - 2\cos x} \right| + k} $ we have,
$ \Rightarrow a = 2$
So this is the required answer.
Hence option (d) is the correct answer.

Note: Given integral
$\int {\dfrac{{5\tan x}}{{\tan x - 2}}dx = x + a\ln \left| {\sin x - 2\cos x} \right| + k} $
Consider the LHS of the above equation we have,
\[ \Rightarrow \int {\dfrac{{5\tan x}}{{\tan x - 2}}dx} \]
Let, \[I = \int {\dfrac{{5\tan x}}{{\tan x - 2}}dx} \]
Now as we know that tan x = (sin x/cos x) so use this property in the above equation we have,
Let, \[I = \int {\dfrac{{5\left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{\dfrac{{\sin x}}{{\cos x}} - 2}}dx} \]
Now simplify it we have,
\[ \Rightarrow I = \int {\dfrac{{5\sin x}}{{\sin x - 2\cos x}}dx} \]
Now 5 sin x is written as, 5 sin x = {(sin x – 2 cos x) + 2 (cos x + 2sin x)} so apply this in the above integral we have,
\[ \Rightarrow I = \int {\dfrac{{\left\{ {\left( {\sin x - 2\cos x} \right) + 2\left( {\cos x + 2\sin x} \right)} \right\}}}{{\sin x - 2\cos x}}dx} \]
Now separate the integral we have,
\[ \Rightarrow I = \int {\dfrac{{\sin x - 2\cos x}}{{\sin x - 2\cos x}}dx} + 2\int {\dfrac{{\cos x + 2\sin x}}{{\sin x - 2\cos x}}dx} \]
Now simplify it we have,
\[ \Rightarrow I = \int {1dx} + 2\int {\dfrac{{\cos x + 2\sin x}}{{\sin x - 2\cos x}}dx} \].................. (1)
Now let, $\sin x - 2\cos x = t$................ (2)
Now differentiate equation (2) w.r.t x we have,
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sin x - 2\cos x} \right) = \dfrac{{dt}}{{dx}}$
Now as we know that $\dfrac{d}{{dx}}\sin x = \cos x,\dfrac{d}{{dx}}\cos x = - \sin x$ so we have,
$ \Rightarrow \left( {\cos x + 2\sin x} \right) = \dfrac{{dt}}{{dx}}$
$ \Rightarrow \left( {\cos x + 2\sin x} \right)dx = dt$.................... (3)
Now substitute the values from equation (2) and (3) in equation (1) we have,
\[ \Rightarrow I = \int {1dx} + 2\int {\dfrac{1}{t}dt} \]
Now as we know that $\int {\dfrac{1}{x}dx} = \ln \left| x \right| + k,\int {1dx} = x + k$, where k is some arbitrary integration constant so we have,
\[ \Rightarrow I = x + 2\ln \left| t \right| + k\]
Now substitute the value of k in the above equation we have,
\[ \Rightarrow I = x + 2\ln \left| {\sin x - 2\cos x} \right| + k\]
$ \Rightarrow \int {\dfrac{{5\tan x}}{{\tan x - 2}}dx = x + 2\ln \left| {\sin x - 2\cos x} \right| + k} $
Now on comparing with, $\int {\dfrac{{5\tan x}}{{\tan x - 2}}dx = x + a\ln \left| {\sin x - 2\cos x} \right| + k} $ we have,
$ \Rightarrow a = 2$
So this is the required answer.
Hence option (d) is the correct answer.