GATE2009 q25 - Revision history

Arjun Suresh at 11:51, 15 July 2014

2014-07-15T11:51:22Z

Arjun Suresh at 19:58, 14 July 2014

2014-07-14T19:58:46Z

Arjun Suresh at 19:58, 14 July 2014

2014-07-14T19:58:16Z

Arjun Suresh at 19:57, 14 July 2014

2014-07-14T19:57:59Z

Arjun Suresh: 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+..."

2014-07-14T19:57:25Z

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+..."

New page

$\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 $(1-\tan x)/(1+\tan x) = \tan(\pi;/4 - x)$, 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 <b>(D)</b> is correct.
{{Template:FBD}}

[[Category: GATE2009]]
[[Category: Calculus questions]]

← Older revision		Revision as of 11:51, 15 July 2014
Line 18:		Line 18:

	[[Category: GATE2009]]		[[Category: GATE2009]]
−	[[Category: Calculus questions]]	+	[[Category: Calculus questions from GATE]]

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

@@ Line 14: / Line 14: @@
 		$$\int^{\pi/4}_0 (1-tanx)/(1+tanx)\,dx = \int^{\pi/4}_0 tan(&pi;/4 - x)\,dx = -\left[\ln \sec(\pi/4 - x)\right]^{\pi/4}_0 =
 /2 \ln 2$$
-		So option <b>(D)</b> is correct.
+		So, option <b>(D)</b> is correct.
 {{Template:FBD}}
 [[Category: GATE2009]]
 [[Category: Calculus questions]]