let give A=∫∫_(0≤y≤x≤1) ((dxdxy)/((1+x^2 )(1+y^2 ))) and B= ∫_0 ^(π/4) ((ln(2cos^2 θ))/(2cos(2θ)))dθ calculate A and prove that B=A.


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Question Number 27691 by abdo imad last updated on 12/Jan/18

$l e t g i v e A = \int \int_{0 ⩽ y ⩽ x ⩽ 1} \frac{d x d x y}{(1 + x^{2}) (1 + y^{2})} a n d$ $B = \int_{0}^{\frac{π}{4}} \frac{l n (2 {c o s}^{2} θ)}{2 c o s (2 θ)} d θ c a l c u l a t e A a n d p r o v e t h a t B = A .$
Commented by abdo imad last updated on 19/Jan/18

$A = \int_{0}^{1} (\int_{0}^{x} \frac{d y}{1 + y^{2}}) \frac{d x}{1 + x^{2}} = \int_{0}^{1} \frac{a r c t a n x}{1 + x^{2}} d x b y p a r t s$ $A = {[a c t a n x . a r c t a n x]}_{0}^{1} - \int_{0}^{1} a r c t a n x \frac{d x}{1 + x^{2}}$ $2 A = \frac{π^{2}}{16} \Rightarrow A = \frac{π^{2}}{32}$ $B = \int_{0}^{\frac{π}{4}} \frac{l n (2 \frac{1 + c o s (2 θ)}{2})}{2 c o s (2 θ)} d θ = \int_{0}^{\frac{π}{4}} \frac{l n (1 + c o s (2 θ))}{2 c o s (2 θ)} d θ$ $= \int_{0}^{\frac{π}{2}} \frac{l n (1 + c o s t)}{2 c o s t} \frac{d t}{2} = \frac{1}{4} \int_{0}^{\frac{π}{2}} \frac{l n (1 + c o s t)}{c o s t} d t l e t i n t r o d u c e$ $F (x) = \int_{0}^{\frac{π}{2}} \frac{l n (1 + x c o s t)}{c o s t} d t w i t h - 1 < x < 1$ $F^{^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x c o s t} t h e c h . t a n (\frac{t}{2}) = u g i v e$ $F^{} (x) = \int_{0}^{1} \frac{1}{1 + x \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \int_{0}^{1} \frac{2 d u}{1 + u^{2} + x - {x u}^{2}}$ $= \int_{0}^{1} \frac{2 d u}{(1 - x) u^{2} + 1 + x} = \frac{2}{1 - x} \int_{0}^{1} \frac{d u}{u^{2} + \frac{1 + x}{1 - x}} t h e c h .$ $u = \sqrt{\frac{1 + x}{1 - x}} t g i v e$ $F^{} (x) = \frac{2}{1 - x} \int_{0}^{\frac{\sqrt{1 - x}}{\sqrt{1 + x}}} \frac{1}{\frac{1 + x}{1 - x} (1 + u^{2})} \frac{\sqrt{1 + x}}{\sqrt{1 - x}} d t$ $= \frac{2}{1 - x} \frac{1 - x}{1 + x} \frac{\sqrt{1 + x}}{\sqrt{1 - x}} \int_{0}^{\frac{\sqrt{1 - x}}{\sqrt{1 + x}}} \frac{d t}{1 + t^{2}}$ $= \frac{2}{\sqrt{1 - x^{2}}} a r c t a n (\sqrt{\frac{1 - x}{1 + x}}) b u t w i t h x = c o s θ$ $a r c t a n (\sqrt{\frac{1 - x}{1 + x}}) = a r c t a n (t a n (\frac{θ}{2})) = \frac{θ}{2}$ $F^{} (x) = \frac{2}{\sqrt{1 - x^{2}}} \frac{a r c o s x}{2} = \frac{a r c o s x}{\sqrt{1 - x^{2}}}$ $F (x) = \int_{0}^{x} \frac{a r c o s t}{\sqrt{1 - t^{2}}} d t + λ (λ = F (0) = 0)$ $\Rightarrow F (x) = \int_{0}^{x} \frac{a r c o s t}{\sqrt{1 - t^{2}}} d t$ $B = \frac{1}{4} F (1) = \frac{1}{4} \int_{0}^{1} \frac{a r c o s t}{\sqrt{1 - t^{2}}} d t b y p a r t s$ $l e t p u t I = \int_{0}^{1} \frac{a r c o s t}{\sqrt{1 - t^{2}}} d t b y p a r t s$ ${I = - a r c o s t . a r c o s t]}_{0}^{1} - \int_{0}^{1} (- a r c o s t) \frac{- d t}{\sqrt{1 - t^{2}}}$ $2 I = \frac{π^{2}}{4} \Rightarrow I = \frac{π^{2}}{8} \Rightarrow F (1) = \frac{π^{2}}{8} a n d B = \frac{π^{2}}{32} = A .$ $f o r F (x) = \int_{0}^{x} \frac{a r c o s t}{\sqrt{1 - t^{2}}} d t l e t i n t e g r a t e b y p a r t s$ ${F (x) = - a r c o t a r c o s t]}_{0}^{x} - \int_{0}^{x} (- a r c o s x) \frac{- d t}{\sqrt{1 - x^{2}}}$ $= \frac{π^{2}}{4} - {(a r c o s x)}^{2} - F (x)$ $\Rightarrow F (x) = \frac{π^{2}}{8} - \frac{1}{2} {(a r c o s x)}^{2} s o$ $\int_{0}^{\frac{π}{2}} \frac{l n (1 + x c o s t)}{c o s t} d t = \frac{π^{2}}{8} - \frac{1}{2} {(a r c o s x)}^{2} .$
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