GATE CSE 2009 | Question: 25

Question

$\int^{\pi/4}_0 (1-\tan x)/(1+\tan x)\,dx $

• A. $0$
• B. $1$  
• C. $\ln 2$
• D. $1/2 \ln 2$

Answer 1

Let $\displaystyle I = \int_{0}^{\frac{\pi}{4}}\frac{1-\tan x}{1+\tan x}dx =  \int_{0}^{\frac{\pi}{4}}\frac{\cos x-\sin x}{\cos x+\sin x}dx$

Now put $\cos x+\sin x=t\;,$ Then $\left(-\sin x+\cos x\right)dx = dt$ and changing limit

So we get $\displaystyle I = \int_{1}^{\sqrt{2}}\frac{1}{t}dt = \left[\ln t\right] = \ln(\sqrt{2}) = \frac{\ln 2}{2}$

Correct Answer: $D$

Comments

By using tan(45-x) i am getting

-ln⁡2/2

In my answer - sign is also there.

Please correct me..

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I m also getting minus sign..

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No. I am not getting any minus sign even doing it by tan(pi/4-x) .

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yes , sandeep , u r correct.. I did  some mistake.. now got it..

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no minus sign would not come

Answer 2

Here’s a simpler method done by using the formula : tan (a – b)

Answer 3

$\int_{0}^{\frac{\pi}{4}} \dfrac{1 - \tan x}{1 + \tan x}\\ \\ = \int_{0}^{\frac{\pi}{4}} \dfrac{\cos x - \sin x}{\cos x + \sin x}\\ \text{ Multiply and divide by cos(x)-sin(x)}\\ = \int_{0}^{\frac{\pi}{4}} \dfrac{1-2\cos x\sin x}{\cos 2x}\\ \int_{0}^{\frac{\pi}{4}} \dfrac{1 - \tan x}{1 + \tan x}\\ \\ = \int_{0}^{\frac{\pi}{4}} \dfrac{\cos x - \sin x}{\cos x + \sin x}$
 

Answer is D.

Comments

Step jump ??