Question

Find the value of $\int {\dfrac{{1 - \tan x}}{{1 + \tan x}}dx}$
A $\log \cos (\dfrac{\pi }{4} - x)$
B $\log \cos (\dfrac{\pi }{4} + x)$
C $\log \sin (\dfrac{\pi }{4} - x)$
D $\log \sin (\dfrac{\pi }{4} + x)$

Answer 1

Here we have to integrate $\dfrac{{1 - \tan x}}{{1 + \tan x}}$with respect to x which can be done by using the identity of $\tan \left( {\dfrac{\pi }{4} - x} \right)$ to simplify the term $\dfrac{{1 - \tan x}}{{1 + \tan x}}$given in the question and simplifying the expression and making it easier to solve integrate and further solve it to get the solution of the question. Let us see the complete step by step solution to understand how to use the identity of the$\tan \left( {\dfrac{\pi }{4} - x} \right)$ to simplify the expression of the question and it’s solving process.

Complete step by step solution:
Here we have been asked to integrate the expression $\dfrac{{1 - \tan x}}{{1 + \tan x}}$ with respect to x and let I be the integral of the expression $\dfrac{{1 - \tan x}}{{1 + \tan x}}$that is-
$I = \int {\dfrac{{1 - \tan x}}{{1 + \tan x}}dx}$
So first of all we would first simplify the expression$\dfrac{{1 - \tan x}}{{1 + \tan x}}$ for which we would use the identity of $\tan \left( {\dfrac{\pi }{4} - x} \right)$ that identity is -
\tan \left( {\dfrac{\pi }{4} - x} \right) = \dfrac{{\tan \left\\{ {\dfrac{\pi }{4}} \right\\} - \tan x}}{{1 + \tan \left\\{ {\dfrac{\pi }{4}} \right\\}\tan x}}
$\tan \left( {\dfrac{\pi }{4} - x} \right) = \dfrac{{1 - \tan x}}{{1 + \tan x}}$
Now from above equation we can get a relation between $\dfrac{{1 - \tan x}}{{1 + \tan x}}$and $\tan \left( {\dfrac{\pi }{4} - x} \right)$using this relation and substituting the value of $\dfrac{{1 - \tan x}}{{1 + \tan x}}$as $\tan \left( {\dfrac{\pi }{4} - x} \right)$in the expression given in the question that is $\int {\dfrac{{1 - \tan x}}{{1 + \tan x}}dx}$ we get –
$\int {\tan \left( {\dfrac{\pi }{4} - x} \right)}$
Now I = \int {\tan \left( {\dfrac{\pi }{4} - x} \right)} $$$$dx
Here we would to be using a simple identity of integration to further solve the question and get the desired result that is –
$\int {\tan \left( x \right)dx = - \log \sec x = \log \cos x}$
Now using the above identity to solve the I = \int {\tan \left( {\dfrac{\pi }{4} - x} \right)} $$$$dxand it becomes$I = \log \cos (\dfrac{\pi }{4} - x) + C$
Where C being the integrating constant
So the resultant answer is $I = \log \cos (\dfrac{\pi }{4} - x) + C$
Now the resultant answer is equal to the option A from the options mentioned above that is –
$\log \cos (\dfrac{\pi }{4} - x)$

So, the correct option is the option A.

Note:
While solving the above question the usage of correct identity at the correct place should be taken into the account also there is another method to solve this question in that method the main substitution is of the $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and simplifying the expression and later on integrating that expression and simplifying it further to get the right answer but this method is a bit time consuming also .