calculate by two methods ∫_0 ^1 ∫_0 ^(π/2) ((dx dt)/(1+x^2 tan^2 t)) then find the value of ∫


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Question Number 31083 by abdo imad last updated on 02/Mar/18

$c a l c u l a t e b y t w o m e t h o d s \int_{0}^{1} \int_{0}^{\frac{π}{2}} \frac{d x d t}{1 + x^{2} {t a n}^{2} t}$ $t h e n f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} t c o t a n t d t .$
Commented by abdo imad last updated on 11/Mar/18

$l e t p u t I = \int_{0}^{\frac{π}{2}} (\int_{0}^{1} \frac{d x}{1 + x^{2} {t a n}^{2} t}) d t c h . x t a n t = u g i v e$ $\int_{0}^{1} \frac{d x}{1 + x^{2} {t a n}^{2} t} = \int_{0}^{1} \frac{1}{1 + u^{2}} \frac{d u}{t a n t}$ $= \frac{1}{t a n t} {[a r c t a n u]}_{0}^{t a n t} = \frac{t}{t a n t} \Rightarrow I = \int_{0}^{\frac{π}{2}} t c o t a n t d t$ $f r o m a n o t h e r s i d e b y f u b i n i t h e o r e m w e h a v e$ $I = \int_{0}^{1} (\int_{0}^{\frac{π}{2}} \frac{d t}{1 + x^{2} {t a n}^{2} t}) d x t h e c h . t a n t = u g i v e$ $\int_{0}^{\frac{π}{2}} \frac{d t}{1 + x^{2} {t a n}^{2} t} = \int_{0}^{\infty} \frac{d u}{(1 + u^{2}) (1 + x^{2} u^{2})} l e t$ $d e c o m p o s e F (u) = \frac{1}{(1 + u^{2}) (1 + x^{2} u^{2})} = \frac{a u + b}{1 + u^{2}} + \frac{c u + d}{1 + x^{2} u^{2}}$ $F (- u) = F (u) \Rightarrow \frac{- a u + b}{1 + u^{2}} + \frac{- c u + d}{1 + x^{2} u^{2}} = F (u) \Rightarrow a = c = 0 \Rightarrow$ $F (u) = \frac{b}{1 + u^{2}} + \frac{d}{1 + x^{2} u^{2}} {l i m}_{u \to \infty} u^{2} F (u) = 0 = b + \frac{d}{x^{2}} \Rightarrow$ $d = - {b x}^{2} \Rightarrow F (u) = \frac{b}{1 + u^{2}} - \frac{{b x}^{2}}{1 + x^{2} u^{2}} w e l o o k t h a t b = \frac{1}{1 - x^{2}}$ $a n d F (u) = \frac{1}{1 - x^{2}} (\frac{1}{1 + u^{2}} - \frac{x^{2}}{1 + x^{2} u^{2}})$ $\int_{0}^{\infty} F (u) d u = \frac{1}{1 - x^{2}} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} - \frac{x^{2}}{1 - x^{2}} \int_{0}^{+ \infty} \frac{d u}{1 + x^{2} u^{2}}$ $= \frac{π}{2 (1 - x^{2})} - \frac{x^{2}}{1 - x^{2}} \int_{0}^{\infty} \frac{1}{1 + α^{2}} \frac{d α}{x} (c h . x u = α)$ $= \frac{π}{2 (1 - x^{2})} - \frac{π x}{2 (1 - x^{2})} = \frac{π (1 - x)}{2 (1 - x) (1 + x)} = \frac{π}{2 (1 + x)}$ $I = \int_{0}^{1} \frac{π d x}{2 (1 + x)} = \frac{π}{2} \int_{0}^{1} \frac{d x}{1 + x} = \frac{π}{2} l n (2) \Rightarrow$ $\int_{0}^{\frac{π}{2}} t c o t a n t d t = \frac{π}{2} l n (2) .$
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