Arjun Suresh (talk | contribs) (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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| − | (C) | + | (C) ln 2 |
| − | '''(D) | + | '''(D) 1/2ln 2''' |
==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ||
$\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.
$\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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