let f(x) =∫_0 ^1 ln(1−xt^3 )dt with ∣x∣<1 1) find a explicit form of f(x) 2)calculate ∫_0 ^1 ln(1−(1/(√2))t^3 )dt 3) calculate A(θ) =∫


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 61978 by maxmathsup by imad last updated on 13/Jun/19

$l e t f (x) = \int_{0}^{1} l n (1 - {x t}^{3}) d t w i t h ∣ x ∣< 1$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{1} l n (1 - \frac{1}{\sqrt{2}} t^{3}) d t$ $3) c a l c u l a t e A (θ) = \int_{0}^{1} l n (1 - s i n θ t^{3}) d t w i t h 0 < θ < \frac{π}{2}$
Commented bymaxmathsup by imad last updated on 14/Jun/19
$1) we have f^′ (x) =−∫_0 ^1 (t^3 /(1−xt^3 ))dt =(1/x) ∫_0 ^1 ((1−xt^3 −1)/(1−xt^3 )) dt =(1/x) −(1/x)∫_0 ^1 (dt/(1−xt^3 )) let α =^3 (√x) ⇒ ∫_0 ^1 (dt/(1−xt^3 )) =∫_0 ^1 (dt/(1−(αt)^3 )) =_(αt =u) ∫_0 ^α (du/(α(1−u^3 ))) =−(1/α) ∫_0 ^α (du/(u^3 −1)) let decompose F(u) =(1/(u^3 −1)) =(1/((u−1)(u^2 +u+1))) =(a/(u−1)) +((bu +c)/(u^2 +u+1)) a =lim_(u→1) (u−1)F(u) =(1/3) lim_(u→+∞) uF(u) =0 =a+b ⇒b =−(1/3) ⇒F(u)=(1/(3(u−1))) +((−(1/3)u +c)/(u^2 +u+1)) F(0) =−1 =−(1/3) +c ⇒c =−1+(1/3) =−(2/3) ⇒F(u)=(1/(3(u−1))) −(1/3) ((u +2)/(u^2 +u +1)) ⇒ ∫_0 ^α (du/(u^3 −1)) =∫_0 ^α (du/(3(u−1))) −(1/6) ∫_0 ^α ((2u +1+3)/(u^2 +u+1)) = [(1/6)ln((((u−1)^2 )/(u^2 +u+1)))]_0 ^α −(1/2) ∫_0 ^α (du/(u^2 +u +1)) ∫_0 ^α (du/(u^2 +u +1)) =∫_0 ^α (du/((u+(1/2))^2 +(3/4))) =_(u+(1/2)=((√3)/2) z) (4/3) ∫_(1/(√3)) ^((2α+1)/(√3)) (1/(1+z^2 )) ((√3)/2)dz =(2/(√3)){ arctan(((2α +1)/(√3)))−arctan((1/(√3))) ⇒ ∫_0 ^α (du/(u^3 −1)) =(1/6)ln(((α^2 −2α +1)/(α^2 +α +1)))−(1/(√3)) arctan(((2α+1)/(√3)))+(1/2) arctan((1/(√3))) ⇒ f^′ (x) =(1/x) +(1/(xα)){(1/6)ln(((α^2 −2α +1)/(α^2 +α +1)))−(1/(√3)) arctan(((2α+1)/(√3)))+(1/2)arctan((1/(√3)))} =A(x) and α =^3 (√x) ⇒ f(x) =∫ A(x)dx +C ....be continued...$
$1) w e h a v e f^{^{'}} (x) = - \int_{0}^{1} \frac{t^{3}}{1 - {x t}^{3}} d t = \frac{1}{x} \int_{0}^{1} \frac{1 - {x t}^{3} - 1}{1 - {x t}^{3}} d t$ $= \frac{1}{x} - \frac{1}{x} \int_{0}^{1} \frac{d t}{1 - {x t}^{3}} l e t α =^{3} \sqrt{x} \Rightarrow \int_{0}^{1} \frac{d t}{1 - {x t}^{3}} = \int_{0}^{1} \frac{d t}{1 - {(α t)}^{3}}$ $=_{α t = u} \int_{0}^{α} \frac{d u}{α (1 - u^{3})} = - \frac{1}{α} \int_{0}^{α} \frac{d u}{u^{3} - 1} l e t d e c o m p o s e$ $F (u) = \frac{1}{u^{3} - 1} = \frac{1}{(u - 1) (u^{2} + u + 1)} = \frac{a}{u - 1} + \frac{b u + c}{u^{2} + u + 1}$ $a = {l i m}_{u \to 1} (u - 1) F (u) = \frac{1}{3}$ ${l i m}_{u \to + \infty} u F (u) = 0 = a + b \Rightarrow b = - \frac{1}{3} \Rightarrow F (u) = \frac{1}{3 (u - 1)} + \frac{- \frac{1}{3} u + c}{u^{2} + u + 1}$ $F (0) = - 1 = - \frac{1}{3} + c \Rightarrow c = - 1 + \frac{1}{3} = - \frac{2}{3} \Rightarrow F (u) = \frac{1}{3 (u - 1)} - \frac{1}{3} \frac{u + 2}{u^{2} + u + 1} \Rightarrow$ $\int_{0}^{α} \frac{d u}{u^{3} - 1} = \int_{0}^{α} \frac{d u}{3 (u - 1)} - \frac{1}{6} \int_{0}^{α} \frac{2 u + 1 + 3}{u^{2} + u + 1}$ $= {[\frac{1}{6} l n (\frac{{(u - 1)}^{2}}{u^{2} + u + 1})]}_{0}^{α} - \frac{1}{2} \int_{0}^{α} \frac{d u}{u^{2} + u + 1}$ $\int_{0}^{α} \frac{d u}{u^{2} + u + 1} = \int_{0}^{α} \frac{d u}{{(u + \frac{1}{2})}^{2} + \frac{3}{4}} =_{u + \frac{1}{2} = \frac{\sqrt{3}}{2} z} \frac{4}{3} \int_{\frac{1}{\sqrt{3}}}^{\frac{2 α + 1}{\sqrt{3}}} \frac{1}{1 + z^{2}} \frac{\sqrt{3}}{2} d z$ $= \frac{2}{\sqrt{3}} {a r c t a n (\frac{2 α + 1}{\sqrt{3}}) - a r c t a n (\frac{1}{\sqrt{3}}) \Rightarrow$ $\int_{0}^{α} \frac{d u}{u^{3} - 1} = \frac{1}{6} l n (\frac{α^{2} - 2 α + 1}{α^{2} + α + 1}) - \frac{1}{\sqrt{3}} a r c t a n (\frac{2 α + 1}{\sqrt{3}}) + \frac{1}{2} a r c t a n (\frac{1}{\sqrt{3}}) \Rightarrow$ $f^{^{'}} (x) = \frac{1}{x} + \frac{1}{x α} {\frac{1}{6} l n (\frac{α^{2} - 2 α + 1}{α^{2} + α + 1}) - \frac{1}{\sqrt{3}} a r c t a n (\frac{2 α + 1}{\sqrt{3}}) + \frac{1}{2} a r c t a n (\frac{1}{\sqrt{3}})} = A (x)$ $a n d α =^{3} \sqrt{x} \Rightarrow f (x) = \int A (x) d x + C . . . . b e c o n t i n u e d . . .$
Commented bymaxmathsup by imad last updated on 14/Jun/19
$let use another way let α =^3 (√x) ⇒ f(x) =∫_0 ^1 ln(1−(αt)^3 )dt =∫_0 ^1 ln{(1−αt)(1+αt +α^2 t^2 )}dt =∫_0 ^1 ln(1−αt)dt +∫_0 ^1 ln(α^2 t^2 +αt +1)dt changement 1−αt =u give ∫_0 ^1 ln(1−αt)dt =((−1)/α) ∫_1 ^(1−α) ln(u) du =−(1/α)[ulnu−u]_1 ^(1−α) =−(1/α){ (1−α)ln(1−α)−(1−α) +1} =−(1/α){ (1−α)ln(1−α) +α} =((α−1)/α)ln(1−α)−1 . by parts ∫_0 ^1 ln(α^2 t^2 +αt +1)dt =[t ln(α^2 t^2 +αt +1)]_0 ^1 −∫_0 ^1 t ((2α^2 t +α)/(α^2 t^2 +αt +1)) dt =ln(α^2 +α +1) −∫_0 ^1 ((2α^2 t^2 +αt)/(α^2 t^2 +αt +1)) dt ∫_0 ^1 ((2α^2 t^2 +αt)/(α^2 t^2 +αt +1)) dt =∫_0 ^1 ((2(α^2 t^2 +αt +1)−2αt−2 +αt)/(α^2 t^2 +αt +1)) dt =2 −∫_0 ^1 ((αt +2)/(α^2 t^2 +αt +1)) dt =2−(1/(2α)) ∫_0 ^1 ((2α^2 t +α +3α)/(α^2 t^2 +αt +1)) dt =2−(1/(2α))[ln(α^2 t^2 +αt+1)]_0 ^1 −(3/2) ∫_0 ^1 (dt/(α^2 t^2 +αt +1)) ∫_0 ^1 (dt/(α^2 t^2 +αt +1)) =(1/α^2 ) ∫_0 ^1 (dt/(t^2 +(t/α) +(1/α^2 ))) =(1/α^2 ) ∫_0 ^1 (dt/(t^2 ++((2t)/(2α)) +(1/(4α^2 )) +(1/α^2 ) −(1/(4α^2 )))) =(1/α^2 ) ∫_0 ^1 (dt/((t +(1/(2α)))^2 +(3/α^2 ))) =_(t+(1/(2α))=((√3)/α) x) (1/α^2 ) ∫_(1/(2(√3))) ^((2α+1)/(2(√3))) (α^2 /3) (1/(1+x^2 )) ((√3)/α) dx (we take x>0) =(1/(α(√3))) { arctan(((2α+1)/(2(√3))))−arctan((1/(2(√3))))} ⇒ ∫_0 ^1 ((2α^2 t^2 +αt)/(α^2 t^2 +αt+1)) dt = 2−(1/(2α))ln(α^2 +α +1)−(3/(2α(√3))){ arctan(((2α +1)/(2(√3))))−arctan((1/(2(√3))))} ⇒ f(x) =((α−1)/α)ln(1−α) −1 +ln(α^2 +α +1)−2+(1/(2α))ln(α^2 +α +1) +((√3)/(2α)){ arctan(((2α +1)/(2(√3))))−arctan((1/(2(√3))))} with α =^3 (√x)$
$l e t u s e a n o t h e r w a y l e t α =^{3} \sqrt{x} \Rightarrow$ $f (x) = \int_{0}^{1} l n (1 - {(α t)}^{3}) d t = \int_{0}^{1} l n {(1 - α t) (1 + α t + α^{2} t^{2})} d t$ $= \int_{0}^{1} l n (1 - α t) d t + \int_{0}^{1} l n (α^{2} t^{2} + α t + 1) d t$ $c h a n g e m e n t 1 - α t = u g i v e \int_{0}^{1} l n (1 - α t) d t = \frac{- 1}{α} \int_{1}^{1 - α} l n (u) d u$ $= - \frac{1}{α} {[u l n u - u]}_{1}^{1 - α} = - \frac{1}{α} {(1 - α) l n (1 - α) - (1 - α) + 1}$ $= - \frac{1}{α} {(1 - α) l n (1 - α) + α} = \frac{α - 1}{α} l n (1 - α) - 1 . b y p a r t s$ $\int_{0}^{1} l n (α^{2} t^{2} + α t + 1) d t = {[t l n (α^{2} t^{2} + α t + 1)]}_{0}^{1} - \int_{0}^{1} t \frac{2 α^{2} t + α}{α^{2} t^{2} + α t + 1} d t$ $= l n (α^{2} + α + 1) - \int_{0}^{1} \frac{2 α^{2} t^{2} + α t}{α^{2} t^{2} + α t + 1} d t$ $\int_{0}^{1} \frac{2 α^{2} t^{2} + α t}{α^{2} t^{2} + α t + 1} d t = \int_{0}^{1} \frac{2 (α^{2} t^{2} + α t + 1) - 2 α t - 2 + α t}{α^{2} t^{2} + α t + 1} d t$ $= 2 - \int_{0}^{1} \frac{α t + 2}{α^{2} t^{2} + α t + 1} d t = 2 - \frac{1}{2 α} \int_{0}^{1} \frac{2 α^{2} t + α + 3 α}{α^{2} t^{2} + α t + 1} d t$ $= 2 - \frac{1}{2 α} {[l n (α^{2} t^{2} + α t + 1)]}_{0}^{1} - \frac{3}{2} \int_{0}^{1} \frac{d t}{α^{2} t^{2} + α t + 1}$ $\int_{0}^{1} \frac{d t}{α^{2} t^{2} + α t + 1} = \frac{1}{α^{2}} \int_{0}^{1} \frac{d t}{t^{2} + \frac{t}{α} + \frac{1}{α^{2}}} = \frac{1}{α^{2}} \int_{0}^{1} \frac{d t}{t^{2} + + \frac{2 t}{2 α} + \frac{1}{4 α^{2}} + \frac{1}{α^{2}} - \frac{1}{4 α^{2}}}$ $= \frac{1}{α^{2}} \int_{0}^{1} \frac{d t}{{(t + \frac{1}{2 α})}^{2} + \frac{3}{α^{2}}} =_{t + \frac{1}{2 α} = \frac{\sqrt{3}}{α} x} \frac{1}{α^{2}} \int_{\frac{1}{2 \sqrt{3}}}^{\frac{2 α + 1}{2 \sqrt{3}}} \frac{α^{2}}{3} \frac{1}{1 + x^{2}} \frac{\sqrt{3}}{α} d x (w e t a k e x > 0)$ $= \frac{1}{α \sqrt{3}} {a r c t a n (\frac{2 α + 1}{2 \sqrt{3}}) - a r c t a n (\frac{1}{2 \sqrt{3}})} \Rightarrow$ $\int_{0}^{1} \frac{2 α^{2} t^{2} + α t}{α^{2} t^{2} + α t + 1} d t = 2 - \frac{1}{2 α} l n (α^{2} + α + 1) - \frac{3}{2 α \sqrt{3}} {a r c t a n (\frac{2 α + 1}{2 \sqrt{3}}) - a r c t a n (\frac{1}{2 \sqrt{3}})} \Rightarrow$ $f (x) = \frac{α - 1}{α} l n (1 - α) - 1 + l n (α^{2} + α + 1) - 2 + \frac{1}{2 α} l n (α^{2} + α + 1)$ $+ \frac{\sqrt{3}}{2 α} {a r c t a n (\frac{2 α + 1}{2 \sqrt{3}}) - a r c t a n (\frac{1}{2 \sqrt{3}})} w i t h α =^{3} \sqrt{x}$
Commented bymaxmathsup by imad last updated on 14/Jun/19
$f(x) =(((α−1)ln(1−α))/α) −3 +(1+(1/(2α)))ln(α^2 +α +1) +((√3)/(2α)){ arctan(((2α +1)/(2(√3))))−arctan((1/(2(√3))))} .$
$f (x) = \frac{(α - 1) l n (1 - α)}{α} - 3 + (1 + \frac{1}{2 α}) l n (α^{2} + α + 1)$ $+ \frac{\sqrt{3}}{2 α} {a r c t a n (\frac{2 α + 1}{2 \sqrt{3}}) - a r c t a n (\frac{1}{2 \sqrt{3}})} .$
Commented bymaxmathsup by imad last updated on 14/Jun/19
$2) ∫_0 ^1 ln(1−(1/(√2))t^3 ) =f((1/(√2)) ) we have α =x^(1/3) ⇒ α =(2^(−(1/2)) )^(1/3) =2^(−(1/6)) =(1/((^6 (√2)))) ⇒ f(x) =(((α−1)ln(1−α))/α) −3 +(1+(1/(2α)))ln(α^2 +α +1) +((√3)/(2α)){ arctan(((2α +1)/(2(√3))))−arctan((1/(2(√3))))} .$
$2) \int_{0}^{1} l n (1 - \frac{1}{\sqrt{2}} t^{3}) = f (\frac{1}{\sqrt{2}}) w e h a v e α = x^{\frac{1}{3}} \Rightarrow α = {(2^{- \frac{1}{2}})}^{\frac{1}{3}} = 2^{- \frac{1}{6}} = \frac{1}{(^{6} \sqrt{2})} \Rightarrow$ $f (x) = \frac{(α - 1) l n (1 - α)}{α} - 3 + (1 + \frac{1}{2 α}) l n (α^{2} + α + 1)$ $+ \frac{\sqrt{3}}{2 α} {a r c t a n (\frac{2 α + 1}{2 \sqrt{3}}) - a r c t a n (\frac{1}{2 \sqrt{3}})} .$
Commented bymaxmathsup by imad last updated on 14/Jun/19

$3) \int_{0}^{1} l n (1 - s i n θ t^{3}) d θ = f (s i n θ) w e α =^{3} \sqrt{x} \Rightarrow α =^{3} \sqrt{s i n θ}$ $s o t h e v a l u e o f t h i s i n t e g r a l i s k n o w n .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum