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Revision as of 19:57, 14 July 2014

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


(A) 0

(B) 1

(C) $ln 2$

(D) $1/2ln 2$

Solution by Happy Mittal

We know that $(1-\tan x)/(1+\tan x) = \tan(\pi;/4 - x)$, so $$\int^{\pi/4}_0 (1-tanx)/(1+tanx)\,dx = \int^{\pi/4}_0 tan(π/4 - x)\,dx = -\left[\ln \sec(\pi/4 - x)\right]^{\pi/4}_0 = 1/2 \ln 2$$ So option (D) is correct.



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


(A) 0

(B) 1

(C) $ln 2$

(D) $1/2ln 2$

Solution by Happy Mittal[edit]

We know that $(1-\tan x)/(1+\tan x) = \tan(\pi;/4 - x)$, so $$\int^{\pi/4}_0 (1-tanx)/(1+tanx)\,dx = \int^{\pi/4}_0 tan(π/4 - x)\,dx = -\left[\ln \sec(\pi/4 - x)\right]^{\pi/4}_0 = 1/2 \ln 2$$ So option (D) is correct.



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