find f(x) = ∫_0 ^1 arctan(xt^2 )dt


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 41280 by math khazana by abdo last updated on 04/Aug/18

$f i n d f (x) = \int_{0}^{1} a r c t a n ({x t}^{2}) d t$
Commented by math khazana by abdo last updated on 07/Aug/18

$a d d ∣ x ∣< 1 t o Q .$
	Answered by math khazana by abdo last updated on 07/Aug/18
	$we have f^′ (x)= ∫_0 ^1 (t^2 /(1+x^2 t^4 )) dt = ∫_0 ^1 t^2 (Σ_(n=0) ^∞ (−1)^n x^(2n) t^(4n) )dt = Σ_(n=0) ^∞ (−1)^n x^(2n) ∫_0 ^1 t^(4n+2) dt = Σ_(n=0) ^∞ (((−1)^n x^(2n) )/(4n+3)) if 0<x<1 f^′ (x)= (1/(x(√x)))Σ_(n=0) ^∞ (((−1)^n ((√x))^(4n +3) )/(4n+3)) =(1/(x(√x)))ϕ((√x)) with ϕ(t) =Σ_(n=0) ^∞ (−1)^n (t^(4n+3) /(4n+3)) ϕ^′ (t)= Σ_(n=0) ^∞ (−1)^n t^(4n+2) =t^2 Σ_(n=0) ^∞ (−t^4 )^n = (t^2 /(1+t^4 )) ( we take ∣t∣<1) ⇒ ϕ(t) = ∫ (t^2 /(1+t^4 )) dt let decompose F(t) = (t^2 /(1+t^4 )) F(t) = (t^2 /((t^2 +1)^2 −2t^2 )) =(t^2 /((t^2 −(√2)t +1)(t^2 +(√2)t +1))) = ((at+b)/(t^2 −(√2)t +1)) +((ct +d)/(t^2 +(√2)t +1)) F(−t) =F(t) ⇒((−at +b)/(t^2 +(√2)t +1)) +((−ct +d)/(t^2 −(√2)t +1))=F(t)⇒ c=−a and b=d ⇒ F(t) = ((at +b)/(t^2 −(√2)t +1)) +((−at +b)/(t^2 +(√2)t +1)) F(0) =0 = 2b ⇒b=0 F(1) =(1/2) = (a/(2−(√2))) −(a/(2+(√2))) =(((2+(√2)−2+(√2)))/2))a ⇒2(√2)a =1 ⇒a =(1/(2(√2))) ⇒ F(x) = (1/(2(√2))){ (t/(t^2 −(√2)t +1))−(t/(t^2 +(√2)t +1))}⇒ ϕ(t) =(1/(2(√2))) ∫ ((tdt)/(t^2 −(√2)t +1)) −(1/(2(√2))) ∫ ((t dt)/(t^2 +(√2)t +1)) but ∫ ((t dt)/(t^2 −(√2)t +1)) = (1/2) ∫ ((2t −(√2) +(√2))/(t^2 −(√2)t +1))dt =(1/2)ln(t^2 −(√2)t +1) +(1/(√2)) ∫ (dt/(t^2 −2((√2)/2)t +(1/2) +(1/2))) =(1/2)ln(t^2 −(√2)t +1) +(1/(√2)) ∫ (dt/((t−((√2)/2))^2 +(1/2))) =_(t−((√2)/2) =(1/(√2)) u) (1/2)ln(t^2 −(√2)t +1) + (1/(√2)) ∫ (1/((1/2)(1+u^2 ))) (du/(√2)) =(1/2)ln(t^2 −(√2)t +1) + arctan((((√2)t −2)/2)) also ∫ ((tdt)/(t^2 +(√2)t +1)) =_(t=−x) ∫ ((xdx)/(x^2 −(√2)x +1)) =(1/2)ln(x^2 −(√2)x +1) +arctan((((√2)x−2)/2)) =(1/2)ln(t^2 +(√2)t +1)−arctan((((√2)x +2)/2))⇒ ϕ(g)=(1/(2(√2))){ (1/2)ln(t^2 −(√2)t +1) +arctan((((√2)t−2)/2)) −(1/2)ln(t^2 +(√2)t +1) +arctan((((√2)t+2)/2))} +c ϕ(0) =0=c ⇒ ϕ(t)= (1/(2(√2))){ ln((√((t^2 −(√2)t+1)/(t^2 +(√2)t +1)))) +arctan((t/(√2)) −1) +arctan((t/(√2)) +1)} but f^′ (x)=(1/(x(√x))) ϕ((√x)) ⇒ f^′ (x) =(1/(2x(√(2x)))) { ln(√((x−(√(2x))+1)/(x+(√(2x))+1))) +arctan(((√x)/(√2))−1) +arctan(((√x)/(√2)) +1)} with 0<x<1 ....be continued...$
	$w e h a v e f^{^{'}} (x) = \int_{0}^{1} \frac{t^{2}}{1 + x^{2} t^{4}} d t$ $= \int_{0}^{1} t^{2} (\sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} t^{4 n}) d t$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} \int_{0}^{1} t^{4 n + 2} d t$ $= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{4 n + 3} i f 0 < x < 1$ $f^{^{'}} (x) = \frac{1}{x \sqrt{x}} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(\sqrt{x})}^{4 n + 3}}{4 n + 3} = \frac{1}{x \sqrt{x}} φ (\sqrt{x})$ $w i t h φ (t) = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{t^{4 n + 3}}{4 n + 3}$ $φ^{^{'}} (t) = \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{4 n + 2} = t^{2} \sum_{n = 0}^{\infty} {(- t^{4})}^{n}$ $= \frac{t^{2}}{1 + t^{4}} (w e t a k e ∣ t ∣< 1) \Rightarrow$ $φ (t) = \int \frac{t^{2}}{1 + t^{4}} d t l e t d e c o m p o s e$ $F (t) = \frac{t^{2}}{1 + t^{4}}$ $F (t) = \frac{t^{2}}{{(t^{2} + 1)}^{2} - 2 t^{2}} = \frac{t^{2}}{(t^{2} - \sqrt{2} t + 1) (t^{2} + \sqrt{2} t + 1)}$ $= \frac{a t + b}{t^{2} - \sqrt{2} t + 1} + \frac{c t + d}{t^{2} + \sqrt{2} t + 1}$ $F (- t) = F (t) \Rightarrow \frac{- a t + b}{t^{2} + \sqrt{2} t + 1} + \frac{- c t + d}{t^{2} - \sqrt{2} t + 1} = F (t) \Rightarrow$ $c = - a a n d b = d \Rightarrow$ $F (t) = \frac{a t + b}{t^{2} - \sqrt{2} t + 1} + \frac{- a t + b}{t^{2} + \sqrt{2} t + 1}$ $F (0) = 0 = 2 b \Rightarrow b = 0$ $F (1) = \frac{1}{2} = \frac{a}{2 - \sqrt{2}} - \frac{a}{2 + \sqrt{2}} = (\frac{2 + \sqrt{2} - 2 + \sqrt{2})}{2}) a$ $\Rightarrow 2 \sqrt{2} a = 1 \Rightarrow a = \frac{1}{2 \sqrt{2}} \Rightarrow$ $F (x) = \frac{1}{2 \sqrt{2}} {\frac{t}{t^{2} - \sqrt{2} t + 1} - \frac{t}{t^{2} + \sqrt{2} t + 1}} \Rightarrow$ $φ (t) = \frac{1}{2 \sqrt{2}} \int \frac{t d t}{t^{2} - \sqrt{2} t + 1} - \frac{1}{2 \sqrt{2}} \int \frac{t d t}{t^{2} + \sqrt{2} t + 1} b u t$ $\int \frac{t d t}{t^{2} - \sqrt{2} t + 1} = \frac{1}{2} \int \frac{2 t - \sqrt{2} + \sqrt{2}}{t^{2} - \sqrt{2} t + 1} d t$ $= \frac{1}{2} l n (t^{2} - \sqrt{2} t + 1) + \frac{1}{\sqrt{2}} \int \frac{d t}{t^{2} - 2 \frac{\sqrt{2}}{2} t + \frac{1}{2} + \frac{1}{2}}$ $= \frac{1}{2} l n (t^{2} - \sqrt{2} t + 1) + \frac{1}{\sqrt{2}} \int \frac{d t}{{(t - \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}}$ $=_{t - \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} u} \frac{1}{2} l n (t^{2} - \sqrt{2} t + 1)$ $+ \frac{1}{\sqrt{2}} \int \frac{1}{\frac{1}{2} (1 + u^{2})} \frac{d u}{\sqrt{2}}$ $= \frac{1}{2} l n (t^{2} - \sqrt{2} t + 1) + a r c t a n (\frac{\sqrt{2} t - 2}{2}) a l s o$ $\int \frac{t d t}{t^{2} + \sqrt{2} t + 1} =_{t = - x} \int \frac{x d x}{x^{2} - \sqrt{2} x + 1}$ $= \frac{1}{2} l n (x^{2} - \sqrt{2} x + 1) + a r c t a n (\frac{\sqrt{2} x - 2}{2})$ $= \frac{1}{2} l n (t^{2} + \sqrt{2} t + 1) - a r c t a n (\frac{\sqrt{2} x + 2}{2}) \Rightarrow$ $φ (g) = \frac{1}{2 \sqrt{2}} {\frac{1}{2} l n (t^{2} - \sqrt{2} t + 1) + a r c t a n (\frac{\sqrt{2} t - 2}{2})$ $- \frac{1}{2} l n (t^{2} + \sqrt{2} t + 1) + a r c t a n (\frac{\sqrt{2} t + 2}{2})} + c$ $φ (0) = 0 = c \Rightarrow$ $φ (t) =$ $\frac{1}{2 \sqrt{2}} {l n (\sqrt{\frac{t^{2} - \sqrt{2} t + 1}{t^{2} + \sqrt{2} t + 1}}) + a r c t a n (\frac{t}{\sqrt{2}} - 1)$ $+ a r c t a n (\frac{t}{\sqrt{2}} + 1)} b u t f^{^{'}} (x) = \frac{1}{x \sqrt{x}} φ (\sqrt{x}) \Rightarrow$ $f^{^{'}} (x) = \frac{1}{2 x \sqrt{2 x}} {l n \sqrt{\frac{x - \sqrt{2 x} + 1}{x + \sqrt{2 x} + 1}} + a r c t a n (\frac{\sqrt{x}}{\sqrt{2}} - 1)$ $+ a r c t a n (\frac{\sqrt{x}}{\sqrt{2}} + 1)} w i t h 0 < x < 1 . . . . b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum