find the value of ∫_0 ^(π/4) ln(1+tanx)dx


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Question Number 42391 by abdo.msup.com last updated on 24/Aug/18

$f i n d t h e v a l u e o f \int_{0}^{\frac{π}{4}} l n (1 + t a n x) d x$
Commented by maxmathsup by imad last updated on 25/Aug/18

$l e t A = \int_{0}^{\frac{π}{4}} l n (1 + t a n x) d x c h a n g e m e n t x = \frac{π}{4} - t g i v e$ $A = \int_{0}^{\frac{π}{4}} l n (1 + t a n (\frac{π}{4} - t)) d t = \int_{0}^{\frac{π}{4}} l n (1 + \frac{1 - t a n t}{1 + t a n t}) d t$ $= \int_{0}^{\frac{π}{4}} l n (\frac{2}{1 + t a n t}) d t = \frac{π}{4} l n (2) - \int_{0}^{\frac{π}{4}} l n (1 + t a n t) d t$ $= \frac{π}{4} l n (2) - A \Rightarrow 2 A = \frac{π}{4} l n (2) \Rightarrow A = \frac{π}{8} l n (2) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18
	$I=∫_0 ^(Π/4) ln(1+tanx)dx ∫_0 ^(Π/4) ln{1+tan((Π/4)−x)}dx ∫_0 ^(Π/4) ln{1+((1−tanx)/(1+tanx))}dx ∫_0 ^(Π/4) ln((2/(1+tanx)))dx =∫_0 ^(Π/4) ln2 dx−∫_0 ^(Π/4) ln(1+tanx)dx 2I=ln2∫_0 ^(Π/4) dx I=ln2×((Π/4)/2)=ln2×(Π/8)$
	$I = \int_{0}^{\frac{Π}{4}} l n (1 + t a n x) d x$ $\int_{0}^{\frac{Π}{4}} l n {1 + t a n (\frac{Π}{4} - x)} d x$ $\int_{0}^{\frac{Π}{4}} l n {1 + \frac{1 - t a n x}{1 + t a n x}} d x$ $\int_{0}^{\frac{Π}{4}} l n (\frac{2}{1 + t a n x}) d x$ $= \int_{0}^{\frac{Π}{4}} l n 2 d x - \int_{0}^{\frac{Π}{4}} l n (1 + t a n x) d x$ $2 I = l n 2 \int_{0}^{\frac{Π}{4}} d x$ $I = l n 2 \times \frac{\frac{Π}{4}}{2} = l n 2 \times \frac{Π}{8}$
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