Line 8: Line 8:
 
(C) ln 2
 
(C) ln 2
 
 
'''(D) 1/2ln 2'''
+
'''(D) 1/2 ln 2'''
  
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===

Revision as of 19:58, 14 July 2014

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


(A) 0

(B) 1

(C) ln 2

(D) 1/2 ln 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.



blog comments powered by Disqus

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


(A) 0

(B) 1

(C) ln 2

(D) 1/2 ln 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.



blog comments powered by Disqus