Question

Evaluate $\int {\dfrac{{1 - \tan x}}{{1 + \tan x}}} dx$
A) $\log \cos \left( {\dfrac{\pi }{4} - x} \right)$
B) $\log \cos \left( {\dfrac{\pi }{4} + x} \right)$
C) $\log \sin \left( {\dfrac{\pi }{4} - x} \right)$
D) $\log \sin \left( {\dfrac{\pi }{4} + x} \right)$

Answer 1

Here, we have to find the integral of the given function. First we have to simplify the function using the trigonometric identities. Then we will integrate the function using the formula. Integration is a way of adding slices to find the whole. Integration is the act of bringing together smaller components into a single system that functions as one.

Formula Used:
We will use the following formulas:
Trigonometric Identity: $\tan (A \pm B) = \dfrac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}$ ; ${\sec ^{ - 1}}x = \cos x$
Trigonometric Value: $\tan \dfrac{\pi }{4} = 1$
Integral Formula:$\int {\tan x} dx = - \log \left( {\sec x} \right) + C$
Logarithmic Formula: $a\log b = \log {b^a}$

Complete step by step solution:
We will use the formula of trigonometric identity to get the expression given in question.
Using $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$ formula for$\tan \left( {\dfrac{\pi }{4} - x} \right)$, we have
$\Rightarrow \tan \left( {\dfrac{\pi }{4} - x} \right) = \dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) - \tan x}}{{1 + \tan \dfrac{\pi }{4}\tan x}}$
Since $\tan \dfrac{\pi }{4} = 1$, so we have
$\Rightarrow \tan \left( {\dfrac{\pi }{4} - x} \right) = \dfrac{{1 - \tan x}}{{1 + (1)\tan x}}$
$\Rightarrow \tan \left( {\dfrac{\pi }{4} - x} \right) = \dfrac{{1 - \tan x}}{{1 + \tan x}}$

$\Rightarrow \dfrac{{1 - \tan x}}{{1 + \tan x}} = \tan \left( {\dfrac{\pi }{4} - x} \right)$
Substituting the value of $\dfrac{{1 - \tan x}}{{1 + \tan x}} = \tan \left( {\dfrac{\pi }{4} - x} \right)$ , in , we get
$\Rightarrow \int {\dfrac{{1 - \tan x}}{{1 + \tan x}}} dx = \int {\tan \left( {\dfrac{\pi }{4} - x} \right)} dx$

Using $\int {\tan x} dx = - \log \left( {\sec x} \right) + C$, we have
$\Rightarrow \int {\tan \left( {\dfrac{\pi }{4} - x} \right)} dx = - \log \sec \left( {\dfrac{\pi }{4} - x} \right) + C$
Using Logarithmic Formula $a\log b = \log {b^a}$, we have
$\Rightarrow \int {\tan \left( {\dfrac{\pi }{4} - x} \right)} dx = \log {\sec ^{ - 1}}\left( {\dfrac{\pi }{4} - x} \right) + C$
We know that, ${\sec ^{ - 1}}x = \cos x$, so we have
$\Rightarrow \int {\tan \left( {\dfrac{\pi }{4} - x} \right)} dx = \log \cos \left( {\dfrac{\pi }{4} - x} \right) + C$
Arbitrary Constant is neglected, so we have
$\Rightarrow \int {\tan \left( {\dfrac{\pi }{4} - x} \right)} dx = \log \cos \left( {\dfrac{\pi }{4} - x} \right)$
$\Rightarrow \int {\dfrac{{1 - \tan x}}{{1 + \tan x}}} dx = \log \cos \left( {\dfrac{\pi }{4} - x} \right)$

Therefore, $\int {\dfrac{{1 - \tan x}}{{1 + \tan x}}} dx = \log \cos \left( {\dfrac{\pi }{4} - x} \right)$

Note:
We can find the integral also by using the substitution method.
Rewriting tangent in terms of sine and cosine, we have
$\Rightarrow \int {\dfrac{{1 - \tan x}}{{1 + \tan x}}} dx = \int {\dfrac{{1 - \dfrac{{\sin x}}{{\cos x}}}}{{1 + \dfrac{{\sin x}}{{\cos x}}}}} dx$
By cross-multiplication method, we have
$\Rightarrow \int {\dfrac{{1 - \tan x}}{{1 + \tan x}}} dx = \int {\dfrac{{\dfrac{{\cos x - \sin x}}{{\cos x}}}}{{\dfrac{{\cos x + \sin x}}{{\cos x}}}}} dx$
Cancelling the denominator, we have
$\Rightarrow \int {\dfrac{{1 - \tan x}}{{1 + \tan x}}} dx = \int {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} dx$

Substituting the denominator $u = \cos x + \sin x$ and differentiating the denominator, we have $\dfrac{{du}}{{dx}} = \cos x - \sin x$ , we get
$\Rightarrow \int {\dfrac{{1 - \tan x}}{{1 + \tan x}}} dx = \int {\dfrac{1}{u}} du$
$\Rightarrow \int {\dfrac{{1 - \tan x}}{{1 + \tan x}}} dx = \log u + C$
Rewriting the term, we have
$\Rightarrow \int {\dfrac{{1 - \tan x}}{{1 + \tan x}}} dx = \log (\cos x + \sin x) + C$