(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+...")
 
 
(3 intermediate revisions by the same user not shown)
Line 6: Line 6:
 
(B) 1
 
(B) 1
 
 
(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}} }}===
Line 14: Line 14:
 
$$\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 =  
 
$$\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$$
 
1/2 \ln 2$$
So option <b>(D)</b> is correct.
+
So, option <b>(D)</b> is correct.
 
{{Template:FBD}}
 
{{Template:FBD}}
  
 
[[Category: GATE2009]]
 
[[Category: GATE2009]]
[[Category: Calculus questions]]
+
[[Category: Calculus questions from GATE]]

Latest revision as of 11:51, 15 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) $1/2ln 2$

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.



blog comments powered by Disqus