Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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(C) ln 2 | (C) ln 2 | ||
| − | '''(D) 1/ | + | '''(D) 1/2 ln 2''' |
==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ||
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$$\int^{\pi/4}_0 (1-tanx)/(1+tanx)\,dx = \int^{\pi/4}_0 tan(π/4 - x)\,dx = -\left[\ln \sec(\pi/4 - x)\right]^{\pi/4}_0 = | $$\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]] |
$\int^{\pi/4}_0 (1-tanx)/(1+tanx)\,dx $
(A) 0
(B) 1
(C) ln 2
(D) 1/2 ln 2
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.
$\int^{\pi/4}_0 (1-tanx)/(1+tanx)\,dx $
(A) 0
(B) 1
(C) ln 2
(D) 1/2ln 2
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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