(Created page with "$\int^{\pi/4}_0 (1-tanx)/(1+tanx)\,dx $ (A) 0 (B) 1 (C) $ln 2$ '''(D) $1/2ln 2$''' ==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== We know that $(1-\tan x)/(1+...")
 
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(B) 1
 
(B) 1
 
 
(C) $ln 2$
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(C) ln 2
 
 
'''(D) $1/2ln 2$'''
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'''(D) 1/2ln 2'''
  
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===

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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