let F(x)= ∫_0 ^(π/2) ((arctan(xtant))/(tant)) dt find a simple form of f(x) . 2) find the value of ∫


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Question Number 33884 by math khazana by abdo last updated on 26/Apr/18

$l e t F (x) = \int_{0}^{\frac{π}{2}} \frac{a r c t a n (x t a n t)}{t a n t} d t f i n d a s i m p l e$ $f o r m o f f (x) .$ $2) f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{a r c t a n (2 t a n t)}{t a n t} d t .$
Commented by abdo imad last updated on 28/Apr/18

$w e h a v e \frac{d F}{d x} (x) = \int_{0}^{\frac{π}{2}} \frac{t a n t}{(1 + x^{2} {t a n}^{2} t) t a n t} d t$ $= \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x^{2} {t a n}^{2} t} b u t c h a n g e m e n t t a n t = u g i v e$ $\frac{d F}{d x} = \int_{0}^{\infty} \frac{1}{(1 + x^{2} u^{2})} \frac{d u}{1 + u^{2}} l e t d e c o m p o s e$ $F (u) = \frac{1}{(1 + x^{2} u^{2}) (1 + u^{2})} = \frac{a u + b}{1 + x^{2} u^{2}} + \frac{c u + d}{1 + u^{2}}$ $F (- u) = F (u) \Leftrightarrow \frac{- a u + b}{1 + x^{2} u^{2}} + \frac{- c u + d}{1 + u^{2}} = \frac{a u + b}{1 + x^{2} u^{2}} + \frac{c u + d}{1 + u^{2}} \Rightarrow$ $a = 0 a n d c = 0 \Rightarrow F (u) = \frac{b}{1 + x^{2} u^{2}} + \frac{d}{1 + u^{2}}$ ${l i m}_{x \to + \infty} u^{2} F (u) = 0 = \frac{b}{x^{2}} + d \Rightarrow b + {d x}^{2} = 0 \Rightarrow b = - {d x}^{2}$ $F (u) = \frac{- {d x}^{2}}{1 + x^{2} u^{2}} + \frac{d}{1 + u^{2}} w e h a v e F (0) = 1 = - {d x}^{2} + d$ $= (1 - x^{2}) d \Rightarrow d = \frac{1}{1 - x^{2}} i f x^{2} \neq 1 \Rightarrow$ $F (u) = \frac{- x^{2}}{(1 - x^{2}) (1 + x^{2} u^{2})} + \frac{1}{(1 - x^{2}) (1 + u^{2})}$ $\frac{d F}{d x} {(x)}^{} = \frac{- x^{2}}{1 - x^{2}} \int_{0}^{\infty} \frac{d u}{1 + x^{2} u^{2}} + \frac{1}{1 - x^{2}} \int_{0}^{\infty} \frac{d u}{1 + u^{2}}$ $=_{x u = t} \frac{- x^{2}}{1 - x^{2}} \int_{0}^{\infty} \frac{1}{1 + t^{2}} \frac{d t}{x} + \frac{1}{1 - x^{2}} \frac{π}{2}$ $= \frac{- x}{1 - x^{2}} \frac{π}{2} + \frac{π}{2} \frac{1}{1 - x^{2}} = \frac{π}{2 (1 - x^{2})} (1 - x) = \frac{π}{2 (1 + x)}$ $\Rightarrow F (x) = \frac{π}{2} \int_{0}^{x} l n (1 + t) d t + λ b u t λ = F (0) = 0 \Rightarrow$ $F (x) = \frac{π}{2} \int_{0}^{x} l n (1 + t) d t =_{1 + t = u} \frac{π}{2} \int_{1}^{1 + x} l n (u) d u$ $= \frac{π}{2} {[u l n (u) - u]}_{1}^{1 + x} = \frac{π}{2} ((1 + x) l n (1 + x) - 1 - x + 1)$ $= \frac{π}{2} ((x + 1) l n (x + 1) - x)$ $2) \int_{0}^{\frac{π}{2}} \frac{a r c t a n (2 t a n t)}{t a n t} d t = F (2) = \frac{π}{2} (3 l n (3) - 2)$
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