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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 |
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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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