∫_0 ^(π/4) log (1+tan x)dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 151421 by peter frank last updated on 21/Aug/21

$\int_{0}^{\frac{π}{4}} \log (1 + \tan x) dx$
Commented by puissant last updated on 21/Aug/21

$x = \frac{π}{4} - t \to d x = - d t$ $Q = \int_{\frac{π}{4}}^{0} l n (1 + t a n (\frac{π}{4} - t)) (- d t)$ $= \int_{0}^{\frac{π}{4}} l n (1 + \frac{1 - t a n t}{1 + t a n t}) d t$ $= \int_{0}^{\frac{π}{4}} l n (\frac{1 + t a n t}{1 + t a n t} + \frac{1 - t a n t}{1 + t a n t}) d t$ $\Rightarrow Q = \int_{0}^{\frac{π}{4}} l n (\frac{2}{1 + t a n t}) d t$ $\Rightarrow Q = \int_{0}^{\frac{π}{4}} l n 2 d x - \int_{0}^{\frac{π}{4}} l n (1 + t a n t) d t$ $\Rightarrow 2 Q = \frac{π}{4} l n 2$ $∵∴ Q = \frac{π}{8} l n 2 . . .$
Commented by peter frank last updated on 21/Aug/21

$thank you$
	Answered by peter frank last updated on 21/Aug/21

	$x = \frac{π}{4} - θ$ $dx = - d θ$ $(0, \frac{π}{4}) \Leftrightarrow (\frac{π}{4}, 0)$ $\int_{0}^{\frac{π}{4}} \log [1 + \tan (\frac{π}{4} - θ)] d θ$ $\int_{0}^{\frac{π}{4}} \log [1 + \frac{\tan \frac{π}{4} - \tan θ}{1 + \tan \frac{π}{4} \tan θ}]$ $\int_{0}^{\frac{π}{4}} \log [\frac{2}{1 + \tan θ}] d θ$ $\int_{0}^{\frac{π}{4}} \log 2 d θ - \int_{0}^{\frac{π}{4}} \log (1 + \tan θ)$ $2 I = \log 2 {[θ]}_{0}^{\frac{π}{4}}$ $I = \frac{π}{8} \log 2$
	Answered by mathmax by abdo last updated on 21/Aug/21

	$f (a) = \int_{0}^{\frac{π}{4}} \log (1 + atanx) dx (a > 0) \Rightarrow$ $f^{^{'}} (a) = \int_{0}^{\frac{π}{4}} \frac{tanx}{1 + atanx} = \frac{1}{a} \int_{0}^{\frac{π}{4}} \frac{1 + atanx - 1}{1 + atanx} dx$ $= \frac{π}{4 a} - \frac{1}{a} \int_{0}^{\frac{π}{4}} \frac{dx}{1 + atanx} [we have$ $\int_{0}^{\frac{π}{4}} \frac{dx}{1 + atanx} =_{tanx = t} \int_{0}^{1} \frac{dt}{(1 + t^{2}) (1 + at)}$ $F (t) = \frac{1}{(at + 1) (t^{2} + 1)} = \frac{α}{at + 1} + \frac{mt + n}{t^{2} + 1}$ $α = \frac{1}{\frac{1}{a^{2}} + 1} = \frac{a^{2}}{1 + a^{2}}$ $\lim_{t \to + \infty} tF (t) = \frac{α}{a} + m \Rightarrow m = - \frac{α}{a} = - \frac{a}{1 + a^{2}}$ $\Rightarrow F (t) = \frac{a^{2}}{(1 + a^{2}) (at + 1)} + \frac{- \frac{a}{1 + a^{2}} t + n}{t^{2} + 1}$ $F (0) = \frac{a^{2}}{1 + a^{2}} + n = 1 \Rightarrow n = 1 - \frac{a^{2}}{1 + a^{2}} = \frac{1}{1 + a^{2}} \Rightarrow$ $F (t) = \frac{a^{2}}{(a^{2} + 1) (at + 1)} + \frac{- \frac{a}{a^{2} + 1} t + \frac{1}{1 + a^{2}}}{t^{2} + 1}$ $\Rightarrow \int_{0}^{1} F (t) dt = \frac{a^{2}}{(a^{2} + 1)} \int_{0}^{1} \frac{dt}{at + 1} - \frac{1}{a^{2} + 1} \int_{0}^{1} \frac{at - 1}{t^{2} + 1} dt$ $= \frac{a}{a^{2} + 1} {[\ln (at + 1)]}_{0}^{1} - \frac{a}{2 (a^{2} + 1)} {[\ln (t^{2} + 1)]}_{0}^{1} + \frac{1}{a^{2} + 1} \frac{π}{4}$ $= \frac{aln (1 + a)}{a^{2} + 1} - \frac{\ln 2}{2} \times \frac{a}{a^{2} + 1} + \frac{π}{4 (a^{2} + 1)} \Rightarrow$ $f^{^{'}} (a) = \frac{π}{4 a} - \frac{\ln (1 + a)}{a^{2} + 1} + \frac{\ln 2}{2 (a^{2} + 1)} - \frac{π}{4 a (a^{2} + 1)}$ $f (1) = f (1) - f (0) = \int_{0}^{1} f^{^{'}} (a) da = \int_{0}^{\frac{π}{4}} \log (1 + tant) dt$ $= \int_{0}^{1} \frac{π}{4 a} (1 - \frac{1}{a^{2} + 1}) da - \int_{0}^{1} \frac{\ln (1 + a)}{a^{2} + 1} da + \frac{\ln 2}{2} \int_{0}^{1} \frac{da}{1 + a^{2}}$ $= \frac{π}{4} \int_{0}^{1} \frac{a}{a^{2} + 1} da - \int_{0}^{1} \frac{\ln (1 + a)}{1 + a^{2}} da + \frac{π}{4} \times \frac{\ln 2}{2}$ $= \frac{π}{8} {[\ln (a^{2} + 1)]}_{0}^{1} + \frac{π \ln 2}{8} - \int_{0}^{1} \frac{\ln (1 + a)}{1 + a^{2}} da (a = \tan θ)$ $= \frac{π \ln (2)}{8} + \frac{π \ln 2}{8} - \int_{0}^{\frac{π}{4}} \frac{\ln (1 + \tan θ)}{1 + \tan^{2} θ} (1 + \tan^{2} θ) d θ$ $= \frac{π \ln 2}{4} - I \Rightarrow 2 I = \frac{π \ln 2}{4} \Rightarrow I = \frac{π}{8} \ln (2) \Rightarrow$ $\int_{0}^{\frac{π}{4}} \ln (1 + \tan θ) d θ = \frac{π}{8} \ln (2)$
	Commented by peter frank last updated on 21/Aug/21

	$thank you$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum