Question

If we have an integration as $\int {\dfrac{{\sec x - \tan x}}{{\sqrt {{{\sin }^2}x - \sin x} }}} dx = k\ln |f(x) + \sqrt 2 \sqrt {\tan x(\tan x - \sec x)} | + c$, where $c$ is arbitrary constant and $k$ is a fixed constant, then
(This question has multiple correct options)
A) $k = \sqrt 2 $
B) $k = \dfrac{1}{{\sqrt 2 }}$
C) $f(x) = \tan x - \sec x$
D) $f(x) = \sqrt {\tan x + \sec x} $

Answer 1

Hint: In the given question, we are supposed to find the value of the fixed constant $k$ and the function $f(x)$ as we can assume this by looking at the given options.
We are given the following equation:
$\int {\dfrac{{\sec x - \tan x}}{{\sqrt {{{\sin }^2}x - \sin x} }}} dx = k\ln |f(x) + \sqrt 2 \sqrt {\tan x(\tan x - \sec x)} | + c$
We will firstly try to change the left side of the equation so that the integrand present on the left-side only contains the trigonometric functions $\sec $ and $\tan $ only.
Then, we will solve the integral using a substitution method to obtain an expression that looks similar to the right-hand side of the equation, so that we can compare both expressions to get the value of $k$ and $f(x)$.

Complete step-by-step solution:
Let us take \[I = \int {\dfrac{{\sec x - \tan x}}{{\sqrt {({{\sin }^2}x - \sin x} )}}} dx\].
In order to change the left side of the equation in the terms of the trigonometric functions $\sec $ and $\tan $ only, we are firstly multiplying and dividing the radicand in $I$ in the denominator by ${\cos ^2}x$, as follows:
\[I = \int {\dfrac{{\sec x - \tan x}}{{\sqrt {({{\sin }^2}x - \sin x) \times \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}}} }}} dx\]
The above integral can be written as follows by taking ${\cos ^2}x$ inside the bracket:
\[I = \int {\dfrac{{\sec x - \tan x}}{{\sqrt {(\dfrac{{{{\sin }^2}x - \sin x}}{{{{\cos }^2}x}}) \times {{\cos }^2}} x}}} dx\]
Taking ${\cos ^2}x$, which is present in the numerator of the radicand, outside the radical, we get
\[I = \int {\dfrac{{\sec x - \tan x}}{{\cos x\sqrt {(\dfrac{{{{\sin }^2}x - \sin x}}{{{{\cos }^2}x}})} }}} dx\]
Since, $\dfrac{1}{{\cos x}} = \sec x$, so we can write the above integral as:
\[I = \int {\dfrac{{(\sec x - \tan x)\sec x}}{{\sqrt {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - \dfrac{{\sin x}}{{{{\cos }^2}x}}} }}} dx\]
Now, we can write ${\cos ^2}x = \cos x \times \cos x$, thus the above integral becomes:
$I = \int {\dfrac{{(\sec x - \tan x)\sec x}}{{\sqrt {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - \dfrac{{\sin x}}{{\cos x}} \times \dfrac{1}{{\cos x}}} }}dx} $
We know that $\dfrac{{\sin x}}{{\cos x}} = \tan x$.
So, $\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = {\tan ^2}x$
So, the above integral can be further written as:-
\[I = \int {\dfrac{{(\sec x - \tan x)\sec x}}{{\sqrt {({{\tan }^2}x - \tan x\sec x} }}} dx\]
Now, let $\tan x - \sec x = t$.
Then, on differentiating both sides, we get
${\sec ^2}xdx - \sec x.\tan xdx = dt$
(because the differentiation of $\tan x$ is ${\sec ^2}xdx$ and the differentiation of $\sec x$ is $\sec x\tan xdx$.)
Take $\sec xdx$ common from the LHS, we get:
$(\sec x - \tan x)\sec xdx = dt$
Also, we have that:
$\tan x - \sec x = t$
On squaring both sides, we get:
${(\tan x - \sec x)^2} = {t^2}$
Since, ${(a - b)^2} = {a^2} - 2ab + {b^2}$
therefore, we must have that:
${\tan ^2}x - 2\tan x\sec x + {\sec ^2}x = {\operatorname{t} ^2}$
Now, we know that:
${\sec ^2}x = 1 + {\tan ^2}x$
Therefore, we have that:
${\tan ^2}x - 2\tan x\sec x + 1 + {\tan ^2}x = {\operatorname{t} ^2}$
Now, combining the like terms, that is, combining both ${\tan ^2}x$ present on the LHS of the above equation, we get:
$2{\tan ^2}x - 2\tan x\sec x + 1 = {\operatorname{t} ^2}$
Taking $1$ from LHS to RHS and then, taking out $2$ common from the LHS, we get:
$2({\tan ^2}x - \tan x\sec x) = {\operatorname{t} ^2} - 1$
Taking $2$ from LHS to RHS, we get the above equation as:
$({\tan ^2} - \tan x\sec x) = \dfrac{{{\operatorname{t} ^2} - 1}}{2}$
Thus, the integral $I$ can finally be written as:-
\[I = \int {\dfrac{{dt}}{{\sqrt {\dfrac{{{\operatorname{t} ^2} - 1}}{2}} }}} \]
Taking $2$ out of the integral, we get:
$I = \sqrt 2 \int {\dfrac{{dt}}{{\sqrt {{\operatorname{t} ^2} - 1} }}} $
Now, we know the following integration formula:
$\int {\dfrac{{dx}}{{\sqrt {{x^2} - 1} }} = \ln |x + } \sqrt {{x^2} - 1} | + c$, where $c$ is the constant of integration.
Therefore, by applying this formula on the integral $I$, we get:
$I = \sqrt 2 \ln |t + \sqrt {{t^2} - 1} | + c$
Substituting the value of $t$, which is $\tan x - \sec x$ and the value of ${t^2} - 1$, which is $2({\tan ^2}x - \tan x\sec x)$, we get the integral value $I$ as:
$I = \sqrt 2 \ln |\tan x - \sec x + \sqrt {2({{\tan }^2}x - \tan x\sec x)} | + c$
The above integral can also be written as:
$I = \sqrt 2 \ln |\tan x - \sec x + \sqrt 2 \sqrt {\tan x(\tan x - \sec x)} | + c$
Now, comparing the above value of $I$ with the given value of $I$, that is, with $k\ln |f(x) + \sqrt 2 \sqrt {\tan x(\tan x - \sec x)} | + c$, we get that:
$k = \sqrt 2 $ and
$f(x) = \tan x - \sec x$.
Hence, option A) $k = \sqrt 2 $ and option C) $f(x) = \tan x - \sec x$ are the correct options.

Note: We have used the substitution method to find the value of the integral. Also, to solve such types of questions, we need to remember all the trigonometric identities and integration formulae.
Here, we have changed the integrand so that it only contains the trigonometric functions $\tan $ and $\sec $ only because, on the RHS, we are given the expression that contains the trigonometric functions $\tan $ and $\sec $ only.