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

1. $0$
2. $1$
3. $ln 2$
4. $1/2 ln 2$
edited | 981 views

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

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0
By using tan(45-x) i am getting

-ln⁡2/2

In my answer - sign is also there.

0
I m also getting minus sign..
0
No. I am not getting any minus sign even doing it by tan(pi/4-x) .
+2

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

0
no minus sign would not come

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

0
Step jump ??

+1 vote
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