calculate ∫_0 ^1 arctan(x^2 )dx


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Question Number 34227 by abdo imad last updated on 03/May/18

$c a l c u l a t e \int_{0}^{1} a r c t a n (x^{2}) d x$
Commented by math khazana by abdo last updated on 04/May/18

$l e t p u t I = \int_{0}^{1} a r c t a n (x^{2}) d x b y p a r t s u^{^{'}} = 1 a n d$ $v = a r c t a n (x^{2}) \Rightarrow I = {[x a r c t a n (x^{2})]}_{0}^{1} - \int_{0}^{1} x \frac{2 x}{1 + x^{4}} d x$ $= \frac{π}{4} - 2 \int_{0}^{1} \frac{x^{2}}{1 + x^{4}} d x l e t d e c o m p o s e$ $F (x) = \frac{x^{2}}{1 + x^{4}} = \frac{x^{2}}{{(x^{2} + 1)}^{2} - 2 x^{2}} = \frac{x^{2}}{(x^{2} - \sqrt{2} x + 1) (x^{2} + \sqrt{2} x + 1)}$ $F (x) = \frac{a x + b}{x^{2} - \sqrt{2} x + 1} + \frac{c x + d}{x^{2} + \sqrt{2} x + 1}$ $F (- x) = F (x) \Rightarrow c = - a a n d d = b \Rightarrow$ $F (x) = \frac{a x + b}{x^{2} - \sqrt{2} x + 1} + \frac{- a x + b}{x^{2} + \sqrt{2} x + 1}$ $F (0) = 0 = 2 b \Rightarrow b = 0 \Rightarrow$ $F (x) = \frac{a x}{x^{2} - \sqrt{2} x + 1} - \frac{a x}{x^{2} + \sqrt{2} x + 1}$ $F (1) = \frac{1}{2} = \frac{a}{2 - \sqrt{2}} - \frac{a}{2 + \sqrt{2}} = (\frac{2 + \sqrt{2} - 2 + \sqrt{2}}{2}) a$ $= \sqrt{2} a \Rightarrow a = \frac{1}{2 \sqrt{2}} \Rightarrow$ $F (x) = \frac{1}{2 \sqrt{2}} (\frac{x}{x^{2} - \sqrt{2} x + 1} - \frac{x}{x^{2} + \sqrt{2} x + 1})$ $\int_{0}^{1} \frac{x^{2}}{1 + x^{4}} d x = \frac{1}{2 \sqrt{2}} \int_{0}^{1} \frac{x}{x^{2} - \sqrt{2} x + 1} d x - \frac{1}{2 \sqrt{2}} \int_{0}^{1} \frac{x}{x^{2} + \sqrt{2} x + 1} d x$
Commented by math khazana by abdo last updated on 04/May/18

$\int_{0}^{1} \frac{x}{x^{2} - \sqrt{2} x + 1} d x = \frac{1}{2} \int_{0}^{1} \frac{2 x - \sqrt{2} + \sqrt{2}}{x^{2} - \sqrt{2} x + 1} d x$ $= \frac{1}{2} {[l n (x^{2} - \sqrt{2} x + 1)]}_{0}^{1} + \frac{\sqrt{2}}{2} \int_{0}^{1} \frac{d x}{x^{2} - \sqrt{2} x + 1}$ $= \frac{1}{2} l n (2 - \sqrt{2}) + \frac{\sqrt{2}}{2} \int_{0}^{1} \frac{d x}{x^{2} - \sqrt{2} x + 1} b u t$ $\int_{0}^{1} \frac{d x}{x^{2} - \sqrt{2} x + 1} = \int_{0}^{1} \frac{d x}{x^{2} - 2 \frac{\sqrt{2}}{2} x + \frac{1}{4} + \frac{3}{4}}$ $= \int_{0}^{1} \frac{d x}{{(x - \frac{\sqrt{2}}{2})}^{2} + \frac{3}{4}} =_{x - \frac{\sqrt{2}}{2} = \frac{\sqrt{3}}{2} t} \int_{- \frac{\sqrt{2}}{\sqrt{3}}}^{\frac{2}{\sqrt{3}} (1 - \frac{\sqrt{2}}{2})} \frac{\frac{\sqrt{3}}{2} d t}{\frac{3}{4} (1 + t^{2})}$ $= \frac{4}{3} \frac{\sqrt{3}}{2} \int_{- \frac{\sqrt{2}}{\sqrt{3}}}^{\frac{2}{\sqrt{3}} - \frac{\sqrt{2}}{\sqrt{3}}} \frac{d t}{1 + t^{2}} = \frac{2 \sqrt{3}}{3} (a r c t a n (\frac{2}{\sqrt{3}} - \frac{\sqrt{2}}{\sqrt{3}}) + a r c t a n (\frac{\sqrt{2}}{\sqrt{3}})) \Rightarrow$ $\int_{0}^{1} \frac{x}{x^{2} - \sqrt{2} x + 1} d x = \frac{1}{2} l n (2 - \sqrt{2})$ $+ \frac{2 \sqrt{3}}{3} (a r c t a n (\frac{2}{\sqrt{3}} - \frac{\sqrt{2}}{\sqrt{3}}) + a r c t a n (\frac{\sqrt{2}}{\sqrt{3}}))$ $\int_{0}^{1} \frac{x}{x^{2} + \sqrt{2} x + 1} d x =_{x = - t} \int_{0}^{- 1} \frac{t d t}{t^{2} - \sqrt{2} t + 1}$ $a n d t h i s c a c u l a t e d a b o v e . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/May/18

	$l e t I = \int {t a n}^{- 1} (x^{2}) d x$ $= {t a n}^{- 1} (x^{2}) \int d x - \int [\frac{d}{d x} {t a n}^{- 1} x^{2 .} . \int d x] d x$ $= {x t a n}^{- 1} (x^{2}) - \int \frac{2 x}{1 + x^{4}} . x d x$ $= {x t a n}^{- 1} x - 2 \int \frac{x^{2}}{1 + x^{4}} d x$ $i h a v e a l r e d y s o l v e d \int \frac{x^{2}}{1 + x^{4}} d x$ $p l s c o - r e l a t e$
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