1.

Evaluation of\(\displaystyle\int{\dfrac{1-\tan x}{1+\tan x}}\ dx\)is:1. log (sin x - cos x) + c2. log (sin x - cot x) + c3. log (cos x + sin x) + c4. log (cos x - cot x) + c

Answer» Correct Answer - Option 3 : log (cos x + sin x) + c

Concept:

\(\displaystyle\int \dfrac{dt}{t}\)= log t + c

Calculation:

\(\displaystyle∫{\dfrac{1-\tan x}{1+\tan x}}\ dx\)

Let's convert tan x into sin x and cos x.

\(\Rightarrow \displaystyle\int \dfrac{1-\dfrac{\sin x}{\cos x}}{1+\dfrac{\sin x}{\cos x}}dx\) (∵\(\tan x = \dfrac{\sin x}{\cos x}\) )

\(\Rightarrow \displaystyle\int \dfrac{\dfrac{\cos x-\sin x}{\cos x}}{\dfrac{\cos x + \sin x}{\cos x}}dx\)

\(\Rightarrow \displaystyle\int \dfrac{\cos x-\sin x}{{\cos x + \sin x}}\)dx ----(1)

Let cos x + sin x = t

Differentiating thisequation on both sides,

\(\dfrac{\mathrm{d}(\cos x + \sin x)}{\mathrm{d} x}=\dfrac{\mathrm{d} t}{\mathrm{d} x}\)

(-sinx + cosx)dx = dt

(cos x - sin x)dx = dt

Subsitutingthis value in equation (1),

\(\Rightarrow \displaystyle\int \frac{dt}{t}\) (∵\(\displaystyle\int \dfrac{dt}{t}\)= log t + c)

⇒ log t + c

Subsituting the value of t in this equation,

log (cosx + sin x) + c

Hence,\(\displaystyle∫{\dfrac{1-\tan x}{1+\tan x}}\ dx\)=log (cos x + sin x) + c



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