Answer

Verified

341.1k+ views

**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.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

Using the following information to help you answer class 12 chemistry CBSE

Why should electric field lines never cross each other class 12 physics CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Write the difference between soap and detergent class 10 chemistry CBSE

Give 10 examples of unisexual and bisexual flowers

Differentiate between calcination and roasting class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the difference between anaerobic aerobic respiration class 10 biology CBSE

a Why did Mendel choose pea plants for his experiments class 10 biology CBSE