calculate-arctan-x-1-x-2-dx-

Question Number 48173 by Abdo msup. last updated on 20/Nov/18

c a l c u l a t e \int \frac{a r c t a n (x)}{\sqrt{1 + x^{2}}} d x

Commented by maxmathsup by imad last updated on 26/Nov/18

c h a n g e m e n t x = t a n t g i v e I = \int \frac{t}{\sqrt{1 + {t a n}^{2} t}} (1 + {t a n}^{2} t) d θ

= \int θ t \sqrt{1 + {t a n}^{2} t} d t = \int \frac{t}{c o s t} d t c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e

I = \int \frac{2 a r c t a n (u)}{\frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = 4 \int \frac{a r c t a n (u)}{1 - u^{2}} d u l e t

f (α) = \int \frac{a r c t a n (α u)}{1 - u^{2}} d u \Rightarrow f^{^{'}} (α) = \int \frac{u}{(1 - u^{2}) (1 + α^{2} u^{2})} d u

=_{α u = x} \int \frac{x}{α (1 - \frac{x^{2}}{α^{2}}) (1 + x^{2})} \frac{d x}{α} = \int \frac{x d x}{(α^{2} - x^{2}) (1 + x^{2})} l e t d e c o m p o s e

F (x) = \frac{- x}{(x^{2} - α^{2}) (x^{2} + 1)} \Rightarrow F (x) = \frac{a}{x - α} + \frac{b}{x + α} + \frac{c x + d}{x^{2} + 1} w e h a v e

F (x) = \frac{- x}{(x - α) (x + α) (x^{2} + 1)} \Rightarrow a = \frac{- α}{2 α (α^{2} + 1)} = \frac{- 1}{2 (α^{2} + 1)}

b = \frac{α}{(- 2 α) (1 + α^{2})}

{l i m}_{x \to + \infty} x F (x) = 0 = a + b + c \Rightarrow c = \frac{1}{2 (α^{2} + 1)} + \frac{1}{2 (1 + α^{2})} = \frac{1}{1 + α^{2}}

F (0) = 0 = - \frac{a}{α} + \frac{b}{α} + d \Rightarrow 0 = - a + b + α d = \frac{1}{2 (α^{2} + 1)} - \frac{1}{2 (1 + α^{2})} + α d \Rightarrow d = 0 \Rightarrow

F (x) = \frac{- 1}{2 (1 + α^{2}) (x - α)} - \frac{1}{2 (1 + α^{2}) (x + α)} + \frac{1}{1 + α^{2}} \frac{x}{1 + x^{2}} \Rightarrow

\int F (x) d x = - \frac{1}{2 (1 + α^{2})} l n ∣ x^{2} - α^{2} ∣ + \frac{1}{2 (1 + α^{2})} l n (x^{2} + 1) + c \Rightarrow

f^{^{'}} (α) = - \frac{1}{2 (1 + α^{2})} l n ∣ α^{2} u^{2} - α^{2} ∣ + \frac{1}{2 (1 + α^{2})} l n (α^{2} u^{2} + 1) + c \Rightarrow

f (α) = \frac{1}{2} \int_{.}^{α} \frac{l n ∣ t^{2} u^{2} - t^{2} ∣}{1 + t^{2}} d t + \frac{1}{2} \int \frac{l n (t^{2} u^{2} + 1)}{1 + t^{2}} d t + c α + λ \dots b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 26/Nov/18

l e t f i n d I a t f o r m o f s e r i e w e h a v e I = \int \frac{t}{c o s t} d t

I = \int \frac{t}{\frac{e^{i t} + e^{- i t}}{2}} d t = \int \frac{2 t}{e^{i t} + e^{- i t}} d t = \int \frac{2 t e^{- i t}}{1 + e^{- 2 i t}} d t

= 2 \int t e^{- i t} (\sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- 2 i n t}) d t = 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} \int t e^{- (2 n + 1) i t} d t

= 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} A_{n} b y p a r t s A_{n} = - \frac{1}{(2 n + 1) i} t e^{- (2 n + 1) i t} - \int - \frac{1}{(2 n + 1) i} e^{- (2 n + 1) i t} d t

= \frac{i t}{2 n + 1} e^{- (2 n + 1) i t} - \frac{i}{2 n + 1} (- \frac{1}{(2 n + 1) i}) e^{- (2 n + 1) i t}

= \frac{i t}{2 n + 1} e^{- (2 n + 1) i t} + \frac{1}{{(2 n + 1)}^{2}} e^{- (2 n + 1) i t} \Rightarrow

\sum_{n = 0}^{\infty} A_{n} = i t \sum_{n = 0}^{\infty} \frac{e^{- (2 n + 1) i t}}{2 n + 1} + \sum_{n = 0}^{\infty} \frac{e^{- (2 n + 1) i t}}{{(2 n + 1)}^{2}} \dots . b e c o n t i n u e d \dots

Answered by Abdulhafeez Abu qatada last updated on 21/Nov/18