1) prove the existence of the integral I=∫_0 ^(π/2) ((ln(1+cosx))/(cosx))dx 2)prove that I= ∫∫_D ((siny)/(1+cosx cosy))dxdy with D=[0,(π/2)]^2 3)find the value of I.


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Question Number 27684 by abdo imad last updated on 12/Jan/18

$1) p r o v e t h e e x i s t e n c e o f t h e i n t e g r a l$ $I = \int_{0}^{\frac{π}{2}} \frac{l n (1 + c o s x)}{c o s x} d x$ $2) p r o v e t h a t I = \int \int_{D} \frac{s i n y}{1 + c o s x c o s y} d x d y w i t h$ $D = {[0, \frac{π}{2}]}^{2}$ $3) f i n d t h e v a l u e o f I .$
Commented by abdo imad last updated on 16/Jan/18

$1) t h e c o n v e r g e n c e o f I i s e a s y {l i m}_{x \to \frac{π}{2}} \frac{l n (1 + c o s x)}{c o s x}$ $= {l i m}_{x \to \frac{π}{2}} \frac{- s i n x}{- s i n x (1 + c o s x)} = 1 (b y h o s p i t a l t h e o r e m)$ $s o t h e f u n c t i o n i s l o c a l i n t e g r a b l e i n [0, \frac{π}{2} [$ $2) I = \int \int_{0 ⩽ x ⩽ \frac{π}{2} a n d 0 ⩽ y ⩽ \frac{π}{2}} \frac{s i n y}{1 + c o s x c o s y} d x d y$ $I = \int_{0}^{\frac{π}{2}} (\int_{0}^{\frac{π}{2}} \frac{1}{c o s x} (- (\frac{{1 + c o s x c o s y)}^{,}}{1 + c o s x c o s y}) d y) d x$ $I = \int_{0}^{\frac{π}{2}} \frac{1}{c o s x} {[- l n / 1 + c o s x c o s y /]}_{y = 0}^{y = \frac{π}{2}}] d x$ $I = \int_{0}^{\frac{π}{2}} \frac{l n (1 + c o s x)}{c o s x} d x$ $3) I = \int_{0}^{\frac{π}{2}} (\int_{0}^{\frac{π}{2}} \frac{d x}{1 + c o s y c o s x}) s i n y d y b u t$ $J = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + c o s y c o s x} = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + λ c o s x} (λ = c o s y) a n d t h e c h .$ $t a n (\frac{x}{2}) = t g i v e J = \int_{0^{}}^{1} \frac{\frac{2 d t}{1 + t^{2}}}{1 + λ \frac{1 - t^{2}}{1 + t^{2}}} = \int_{0}^{1} \frac{2 d t}{1 + t^{2} + λ - λ t^{2}}$ $= \int_{0}^{1} \frac{2 d t}{(1 - λ) t^{2} + 1 + λ} = \frac{2}{1 - λ} \int_{0}^{1} \frac{d t}{t^{2} + (\sqrt{\frac{1 + λ}{1 - λ})}} 2$ $= \frac{2}{\sqrt{1 - λ^{2}}} \int_{0}^{\sqrt{\frac{1 - λ}{1 + λ}}} \frac{d u}{1 + u^{2}} (c h . t = \sqrt{\frac{1 + λ}{1 - λ}} u)$ $= \frac{2}{\sqrt{1 - λ^{2}}} a r t a n (\sqrt{\frac{1 - λ}{1 + λ)}}) = \frac{2}{s i n y} a r t a n (t a n (\frac{y}{2)})) = \frac{y}{s i n y}$ $I = \int_{0}^{\frac{π}{2}} \frac{y}{s i n y} s i n y d y = {[\frac{1}{2} y^{2}]}_{0}^{\frac{π}{2}} = \frac{1}{2} \frac{π^{2}}{4} = \frac{π^{2}}{8} .$
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