Question

How do you integrate \[\dfrac{{1 - \tan x}}{{1 + \tan x}}dx\] ?

Answer 1

Hint: We know that the definition of tangent is the ratio of sine to cosine. That is \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] . Applying this to the given problem and taking LCM we simplify it to the simple form. After getting it in the simple form we use substitute rules for further simplification. Also we need to know the integration of certain functions.

Complete step by step solution:
Given, \[\int {\left( {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right)dx} \] .
Now we simplify \[\dfrac{{1 - \tan x}}{{1 + \tan x}}\] by substituting \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] . We get,
 \[\dfrac{{1 - \tan x}}{{1 + \tan x}} = \dfrac{{1 - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{1 + \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}\]
Now taking LCM we have,
 \[ = \dfrac{{\left( {\dfrac{{\cos x - \sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{{\cos x + \sin x}}{{\cos x}}} \right)}}\]
Rearranging the equation we have,
 \[ = \dfrac{{\cos x - \sin x}}{{\cos x}} \times \dfrac{{\cos x}}{{\cos x + \sin x}}\]
Cancelling cosine we have,
 \[ = \dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}\] .
Thus we have,
 \[\int {\left( {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right)dx} = \int {\left( {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right)dx} \]
 \[ = \int {\dfrac{{\left( {\cos x - \sin x} \right)}}{{\left( {\cos x + \sin x} \right)}}dx} \]
Take \[\cos x + \sin x = t\] . Differentiating implicitly with respect to ‘x’ we have,
We know the differentiation of cosine is negative sine and differentiation of sine is cosine.
 \[( - \sin x + \cos x)dx = dt\]
Rearranging the terms in brackets we have,
 \[(\cos x - \sin x)dx = dt\] .
Now substituting this in the above integral we have,
 \[ = \int {\dfrac{1}{t}dt} \]
We know that the integration of \[\dfrac{1}{t}\] is a logarithm function.
 \[ = \log t + c\]
Where ‘c’ is the integration constant.
But we need the answer in terms of ‘x’.
We have \[\cos x + \sin x = t\] , substituting this we have,
 \[ = \log (\cos x + \sin x) + c\] .
Thus we have,
 \[\int {\left( {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right)dx} = = \log (\cos x + \sin x) + c\] , where ‘c’ is the integration constant.
So, the correct answer is “$ \log (\cos x + \sin x) + c$”.

Note: Here we have an indefinite integral that is no upper limit and lower limit. Hence, in the case of indefinite integral we have integration constant. In definite integral we have lower limits and upper limits. Hence, in the case of definite integral we don’t have integration constant. As we can see in the above problem by using the substitute rule we can simplify the problem easily.