let-x-lt-1-calculate-F-x-0-1-ln-1-xt-2-dt-2-find-the-value-of-0-1-ln-1-1-2-t-2-dt-3-find-the-value-of-A-0-1-ln-1-sin-t-2-dt-

Question Number 38458 by maxmathsup by imad last updated on 25/Jun/18

l e t ∣ x ∣< 1 c a l c u l a t e F (x) = \int_{0}^{1} l n (1 + {x t}^{2}) d t

2) f i n d t h e v a l u e o f \int_{0}^{1} l n (1 + \frac{1}{2} t^{2}) d t

3) f i n d t h e v a l u e o f A (θ) = \int_{0}^{1} l n (1 + s i n θ t^{2}) d t .

Commented by math khazana by abdo last updated on 28/Jun/18

w e h a v e F^{^{'}} (x) = \int_{0}^{1} \frac{t^{2}}{1 + {x t}^{2}} d t

= \frac{1}{x} \int_{0}^{1} \frac{{x t}^{2} + 1 - 1}{{x t}^{2} + 1} d t

= \frac{1}{x} - \frac{1}{x} \int_{0}^{1} \frac{d t}{1 + {x t}^{2}} s o i f 0 < x < 1 c h a n g e m e n t \sqrt{x} t = u g i v e

\int_{0}^{1} \frac{d t}{1 + {x t}^{2}} = \int_{0}^{\sqrt{x}} \frac{1}{1 + u^{2}} \frac{d u}{\sqrt{x}}

= \frac{1}{\sqrt{x}} a r c t a n (\sqrt{x}) \Rightarrow

F^{^{'}} (x) = \frac{1}{x} - \frac{a r c t a n (\sqrt{x})}{x \sqrt{x}} \Rightarrow

F (x) = \int_{1}^{x} \frac{d t}{t} - \int_{1}^{x} \frac{a r c t a n (\sqrt{t})}{t \sqrt{t}} d t + c

= l n (x) - \int_{1}^{x} \frac{a r c t a n (\sqrt{t})}{t \sqrt{t}} d t + c b u t c h a n g . \sqrt{t} = u

g i v e \int_{1}^{x} \frac{a r c t a n (\sqrt{t})}{t \sqrt{t}} d t = \int_{1}^{\sqrt{x}} \frac{a r c t a n (u)}{u^{2} . u} 2 u d u

= 2 \int_{1}^{\sqrt{x}} \frac{a r c t a n (u)}{u^{2}} d u b y p a r t s

\int_{1}^{\sqrt{x}} \frac{a r c t a n u}{u^{2}} d u = {[- \frac{1}{u} a r c t a n (u)]}_{1}^{\sqrt{x}}

+ \int_{1}^{\sqrt{x}} \frac{1}{u (1 + u^{2})} d u = \frac{π}{4} - \frac{a r c t a n (\sqrt{x})}{\sqrt{x}}

+ \int_{1}^{\sqrt{x}} (\frac{1}{u} - \frac{u}{1 + u^{2}}) d u

= \frac{π}{4} - \frac{a r c t a n (\sqrt{x})}{\sqrt{x}} + {[l n (\frac{u}{\sqrt{1 + u^{2}}})]}_{1}^{\sqrt{x}}

= \frac{π}{4} - \frac{a r c t a n (\sqrt{x})}{\sqrt{x}} + l n (\frac{\sqrt{x}}{\sqrt{1 + x}}) + l n (\sqrt{2}) \Rightarrow

F (x) = l n (x) - \frac{π}{2} + \frac{2 a r c t a n (\sqrt{x})}{\sqrt{x}} - 2 l n (\frac{\sqrt{x}}{\sqrt{1 + x}})

- l n (2) + c

F (1) = \int_{0}^{1} l n (1 + t^{2}) d t = - \frac{π}{2} + \frac{π}{2} - 2 l n (\frac{1}{\sqrt{2}})

- l n (2) + c = c \Rightarrow

F (x) = - \frac{π}{2} + \frac{2 a r c t a n (\sqrt{x})}{\sqrt{x}} + l n (1 + x) - l n (2)

+ \int_{0}^{1} l n (1 + t^{2}) d t b y p a r t s

\int_{0}^{1} l n (1 + t^{2}) d t = {[t l n (1 + t^{2})]}_{0}^{1} - \int_{0}^{1} \frac{t 2 t}{1 + t^{2}} d t

= l n (2) - 2 \int_{0}^{1} \frac{1 + t^{2} - 1}{1 + t^{2}} d t

= l n (2) - 2 + 2 \int_{0}^{1} \frac{d t}{1 + t^{2}} = l n (2) - 2 + \frac{π}{2} \Rightarrow

F (x) = \frac{2 a r c t a n (\sqrt{x})}{\sqrt{x}} + l n (1 + x) - 2 w i t h 0 < x < 1

Commented by math khazana by abdo last updated on 28/Jun/18

i f - 1 < x < 0 w e h a v e F^{^{'}} (x) = \frac{1}{x} - \frac{1}{x} \int_{0}^{1} \frac{d t}{1 + {x t}^{2}}

\int_{0}^{1} \frac{d t}{1 + {x t}^{2}} = \int_{0}^{1} \frac{d t}{1 - (- x) t^{2}} =_{\sqrt{- x} t = u} \int_{0}^{\sqrt{- x}} \frac{1}{1 - u^{2}} \frac{d u}{\sqrt{- x}}

= \frac{1}{\sqrt{- x}} \int_{0}^{\sqrt{- x}} \frac{d u}{1 - u^{2}} = \frac{1}{2 \sqrt{- x}} \int_{0}^{\sqrt{- x}} (\frac{1}{1 + u} + \frac{1}{1 - u}) d u

= \frac{1}{2 \sqrt{- x}} {[l n (\frac{1 + u}{1 - u})]}_{0}^{\sqrt{- x}} = \frac{1}{2 \sqrt{- x}} l n (\frac{1 + \sqrt{- x}}{1 - \sqrt{- x}}) \Rightarrow

F^{^{'}} (x) = \frac{1}{x} - \frac{1}{2 x \sqrt{- x}} l n (\frac{1 + \sqrt{- x}}{1 - \sqrt{- x}}) \Rightarrow

F (x) = l n (- x) - \int_{1}^{x} \frac{1}{2 t \sqrt{- t}} l n (\frac{1 + \sqrt{- t}}{1 - \sqrt{- t}}) d t + c

\dots b e c o n t i n u e d \dots

Commented by math khazana by abdo last updated on 28/Jun/18

2) \int_{0}^{1} l n (1 + \frac{1}{2} t^{2}) d t = F (\frac{1}{2}) = \frac{2 a r c t a n (\frac{1}{\sqrt{2}})}{\frac{1}{\sqrt{2}}} + l n (\frac{3}{2})

- 2 = 2 \sqrt{2} a r c t a n (\frac{1}{\sqrt{2}}) + l n (3) - l n (2) - 2 .