find-the-value-of-0-1-arctan-1-x-2-dx-

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)