find the value of ∫_0 ^1 arctan((√(1−x^2 )))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 32352 by abdo imad last updated on 23/Mar/18

$f i n d t h e v a l u e o f \int_{0}^{1} a r c t a n (\sqrt{1 - x^{2}}) d x$
Commented by abdo imad last updated on 26/Mar/18

$c h . x = s i n t \Rightarrow I = \int_{0}^{\frac{π}{2}} a r c t a n (c o s t) c o s t d t l e t i n t e g r a t e$ $b y p a r t s u = a r c t a n (c o s t) a n d v^{^{'}} = c o s t$ $I = {[s i n t a r c t a n (c o s t)]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{- s i n t}{1 + {c o s}^{2} t} s i n t d t$ $= \int_{0}^{\frac{π}{2}} \frac{1 - {c o s}^{2} t}{1 + {c o s}^{2} t} d t = \int_{0}^{\frac{π}{2}} \frac{1 - \frac{1 + c o s (2 t)}{2}}{1 + \frac{1 + c o s (2 t)}{2}} d t$ $= \int_{0}^{\frac{π}{2}} \frac{1 - c o s (2 t)}{3 + c o s (2 t)} d t =_{2 t = u} \int_{0}^{π} \frac{1 - c o s u}{3 + c o s u} \frac{d u}{2}$ $= \frac{1}{2} \int_{0}^{π} \frac{1 - c o s u}{3 + c o s u} d u c h . t a n (\frac{u}{2}) = x g i v e$ $I = \frac{1}{2} \int_{0}^{\infty} \frac{1 - \frac{1 - x^{2}}{1 + x^{2}}}{3 + \frac{1 - x^{2}}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}} = \int_{0}^{\infty} \frac{1 + x^{2} - 1 + x^{2}}{(3 + 3 x^{2} + 1 - x^{2}) (1 + x^{2})} d x$ $= \int_{0}^{\infty} \frac{2 x^{2}}{(4 + 2 x^{2}) (1 + x^{2})} d x = \int_{0}^{\infty} \frac{x^{2}}{(x^{2} + 1) (x^{2} + 2)} d x$ $= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{x^{2}}{(x^{2} + 1) (x^{2} + 2)} d x l e t c o n s i d e r$ $φ (z) = \frac{z^{2}}{(z^{2} + 1) (z^{2} + 2)} t h e p o l e s o f φ a r e$ $i, - i, i \sqrt{2}, - i \sqrt{2} a n d a l l a r e s i m p l e s$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (R e s (φ, i) + R e s (φ, i \sqrt{2}))$ $φ (z) = \frac{z^{2}}{(z - i) (z + i) (z - i \sqrt{2}) (z + i \sqrt{2})}$ $R e s (φ, i) = {l i m}_{z \to i} (z - i) φ (z) = \frac{- 1}{(2 i) (1)} = \frac{- 1}{2 i}$ $R e s (φ, i \sqrt{2}) = {l i m}_{z \to i \sqrt{2}} (z - i \sqrt{2}) φ (z) = \frac{- 2}{(- 1) (2 i \sqrt{2})} = \frac{1}{i \sqrt{2}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (\frac{- 1}{2 i} + \frac{1}{i \sqrt{2}}) = - π + \frac{2 π}{\sqrt{2}} = - π + π \sqrt{2}$ $= (\sqrt{2} - 1) π .$
Commented by abdo imad last updated on 26/Mar/18

$I = \frac{1}{2} \int_{- \infty}^{+ \infty} φ (z) d z = \frac{π}{2} (\sqrt{2} - 1) .$
	Answered by sma3l2996 last updated on 25/Mar/18

	$I = \int_{0}^{1} a r c t a n (\sqrt{1 - x^{2}}) d x$ $b y p a r t s$ $u = a r c t a n (\sqrt{1 - x^{2}}) \Rightarrow u^{'} = \frac{- x}{(2 - x^{2}) \sqrt{1 - x^{2}}}$ $v^{'} = 1 \Rightarrow v = x$ $I = {[x a r c t a n (\sqrt{1 - x^{2}})]}_{0}^{1} + \int_{0}^{1} \frac{x^{2}}{(2 - x^{2}) \sqrt{1 - x^{2}}} d x$ $= - \int_{0}^{1} \frac{2 - x^{2} - 2}{(2 - x^{2}) \sqrt{1 - x^{2}}} d x = 2 \int_{0}^{1} \frac{d x}{(2 - x^{2}) \sqrt{1 - x^{2}}} - \int_{0}^{1} \frac{d x}{\sqrt{1 - x^{2}}}$ $l e t x = s i n t \Rightarrow d x = c o s t d t$ $I = 2 \int_{0}^{π / 2} \frac{d t}{2 - {s i n}^{2} t} - \frac{π}{2}$ $u = t a n t \Rightarrow d t = \frac{d u}{1 + u^{2}}$ $2 - {s i n}^{2} t = 2 - \frac{{t a n}^{2} t}{1 + {t a n}^{2} t} = \frac{2 + {t a n}^{2} t}{1 + {t a n}^{2} t} = \frac{2 + u^{2}}{1 + u^{2}}$ $I = 2 \int_{0}^{\infty} \frac{d u}{2 + u^{2}} - \frac{π}{2} = \int_{0}^{\infty} \frac{d u}{1 + {(\frac{u}{\sqrt{2}})}^{2}} - \frac{π}{2} = \sqrt{2} {[a r c t a n (\frac{u}{\sqrt{2}})]}_{0}^{\infty} - \frac{π}{2}$ $I = \frac{π \sqrt{2}}{2} - \frac{π}{2} = \frac{π}{2} (\sqrt{2} - 1)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum