find f(a) =∫_0 ^∞ (dx/(x^3 +a^3 )) with a>0 2)find g(a)=∫_0 ^∞ (dx/((x^3 +a^3 )^2 )) 3)find the value of ∫_0 ^∞ (dx/((1+x^3 )^2 )) 4)find the value of ∫


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Question Number 44202 by abdo.msup.com last updated on 23/Sep/18

$f i n d f (a) = \int_{0}^{\infty} \frac{d x}{x^{3} + a^{3}} w i t h a > 0$ $2) f i n d g (a) = \int_{0}^{\infty} \frac{d x}{{(x^{3} + a^{3})}^{2}}$ $3) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{d x}{{(1 + x^{3})}^{2}}$ $4) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{d x}{8 x^{3} + 1}$
Commented bymaxmathsup by imad last updated on 25/Sep/18

$w e h a v e f (a) = \int_{0}^{\infty} \frac{d x}{x^{3} + a^{3}} \Rightarrow f^{^{'}} (a) = - \int_{0}^{\infty} \frac{3 a^{2}}{{(x^{3} + a^{3})}^{2}} d x \Rightarrow$ $\int_{0}^{\infty} \frac{d x}{{(x^{3} + a^{3})}^{2}} = \frac{- 1}{3 a^{2}} f^{^{'}} (a) b u t f (a) = \frac{2 π}{3 \sqrt{3}} a^{- 2} \Rightarrow f^{^{'}} (a) = - \frac{4 π}{3 \sqrt{3}} a^{- 3} \Rightarrow$ $\int_{0}^{\infty} \frac{d x}{{(x^{3} + a^{3})}^{2}} = \frac{- 1}{3 a^{2}} (- \frac{4 π}{3 \sqrt{3} a^{3}}) = \frac{4 π}{9 a^{5} \sqrt{3}} .$
Commented bymaxmathsup by imad last updated on 25/Sep/18

$l e t t a k e a = 1 \Rightarrow \int_{0}^{\infty} \frac{d x}{{(x^{3} + 1)}^{2}} = \frac{4 π}{9 \sqrt{3}} .$
Commented bymaxmathsup by imad last updated on 25/Sep/18
$1) changement x =at give f(a) = ∫_0 ^∞ ((adt)/(a^3 (1+t^3 ))) =(1/a^2 ) ∫_0 ^∞ (dt/(t^3 +1)) let I =∫_0 ^∞ (dt/(t^3 +1)) ⇒ I =_(t=(1/u)) −∫_0 ^∞ (1/((1/u^3 ) +1)) ((−du)/u^2 ) =∫_0 ^∞ (du/((1/u) +u^2 )) =∫_0 ^∞ (u/(u^(3 ) +1)) du ⇒ 2I = ∫_0 ^∞ (dt/(t^3 +1)) +∫_0 ^∞ (t/(t^3 +1))dt =∫_0 ^∞ ((t+1)/(t^3 +1))dt =∫_0 ^∞ (dt/(t^2 −t +1)) = ∫_0 ^∞ (dt/((t−(1/2))^2 +(3/4))) =_(t−(1/2)=((√3)/2)u) (4/3)∫_(−(1/(√3))) ^(+∞) (1/(1+u^2 )) ((√3)/2)du = (2/(√3)) ∫_(−(1/(√3))) ^(+∞) (du/(1+u^2 )) =(2/(√3)) [arctan(u)]_(−((1 )/(√3))) ^(+∞) =(2/(√3)){(π/2) +(π/6)}=(2/(√3)) .((2π)/3) =((4π)/(3(√3))) ⇒ I = ((2π)/(3(√3))) ⇒f(a) = ((2π)/(3a^2 (√3))) .$
$1) c h a n g e m e n t x = a t g i v e f (a) = \int_{0}^{\infty} \frac{a d t}{a^{3} (1 + t^{3})} = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{d t}{t^{3} + 1}$ $l e t I = \int_{0}^{\infty} \frac{d t}{t^{3} + 1} \Rightarrow I =_{t = \frac{1}{u}} - \int_{0}^{\infty} \frac{1}{\frac{1}{u^{3}} + 1} \frac{- d u}{u^{2}} = \int_{0}^{\infty} \frac{d u}{\frac{1}{u} + u^{2}}$ $= \int_{0}^{\infty} \frac{u}{u^{3} + 1} d u \Rightarrow 2 I = \int_{0}^{\infty} \frac{d t}{t^{3} + 1} + \int_{0}^{\infty} \frac{t}{t^{3} + 1} d t = \int_{0}^{\infty} \frac{t + 1}{t^{3} + 1} d t$ $= \int_{0}^{\infty} \frac{d t}{t^{2} - t + 1} = \int_{0}^{\infty} \frac{d t}{{(t - \frac{1}{2})}^{2} + \frac{3}{4}} =_{t - \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int_{- \frac{1}{\sqrt{3}}}^{+ \infty} \frac{1}{1 + u^{2}} \frac{\sqrt{3}}{2} d u$ $= \frac{2}{\sqrt{3}} \int_{- \frac{1}{\sqrt{3}}}^{+ \infty} \frac{d u}{1 + u^{2}} = \frac{2}{\sqrt{3}} {[a r c t a n (u)]}_{- \frac{1}{\sqrt{3}}}^{+ \infty} = \frac{2}{\sqrt{3}} {\frac{π}{2} + \frac{π}{6}} = \frac{2}{\sqrt{3}} . \frac{2 π}{3} = \frac{4 π}{3 \sqrt{3}}$ $\Rightarrow I = \frac{2 π}{3 \sqrt{3}} \Rightarrow f (a) = \frac{2 π}{3 a^{2} \sqrt{3}} .$
Commented bymaxmathsup by imad last updated on 25/Sep/18

$w e h a v e \int_{0}^{\infty} \frac{d x}{x^{3} + a^{3}} = \frac{2 π}{3 a^{2} \sqrt{3}} l e t t a k e a = \frac{1}{2} \Rightarrow$ $\int_{0}^{\infty} \frac{d x}{x^{3} + \frac{1}{8}} = \frac{2 π}{3 \frac{1}{4} \sqrt{3}} = \frac{8 π}{3 \sqrt{3}} \Rightarrow 8 \int_{0}^{\infty} \frac{d x}{8 x^{3} + 1} = \frac{8 π}{3 \sqrt{3}} \Rightarrow$ $\int_{0}^{\infty} \frac{d x}{8 x^{3} + 1} = \frac{π}{3 \sqrt{3}} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18
	$1)∫_0 ^∞ (dx/(x^3 +a^3 )) x^3 =a^3 tan^2 α 3x^2 dx=a^3 ×2tanαsec^2 αdα dx=((2a^3 )/3)×((tanαsec^2 α)/((a^3 tan^2 α)^(2/3) ))dα=((2a)/3)×((sec^2 α)/(tan^(1/3) α))dα ∫_0 ^(π/2) ((2a)/3)×((sec^2 α)/(tan^(1/3) α))×(1/(a^3 (tan^2 α+1)))×dα (2/(3a^2 ))∫_0 ^(π/2) (dα/(tan^(1/3) α)) (2/(3a^2 ))∫_0 ^(π/2) sin^((−1)/3) αcos^(1/3) αdα using gamma beta function... ∫_0 ^(π/2) sin^(2p−1) αcos^(2q−1) αdα=((⌈(p)⌈q))/(2(⌈p+q))) here 2p−1=((−1)/3) 2p=(2/3) p=(1/3) 2q−1=(1/3) 2q=(4/3) q=(2/3) so ans is (2/(3a^2 ))×((⌈((1/3))×⌈((2/3)))/(2⌈((1/3)+(2/3)))) =(1/(3a^2 ))×((⌈((1/3))×⌈(1−(1/3)))/1)=(1/(3a^2 ))×(π/(sin((π/3))))=((2π)/(3(√3) a^2 )) 2)∫_0 ^∞ (dx/((x^3 +a^3 )^2 ))=∫_0 ^(π/2) ((2a)/3)×((sec^2 α)/(tan^(1/3) α))×(1/({a^3 (tan^2 α+1)}^2 ))dα ∫_0 ^(π/2) ((2a)/3)×((sec^2 α)/(tan^(1/3) α))×(1/(a^6 ×sec^4 α))dα (2/(3a^5 ))∫_0 ^(π/2) ((cos^(1/3) α)/(sin^(1/3) α))×cos^2 αdα (2/(3a^5 ))∫_0 ^(π/2) sin^((−1)/3) αcos^(7/3) αdα 2p−1=((−1)/3) p=(1/3) 2q−1=(7/3) 2q=((10)/3) q=(5/3) (2/(3a^5 ))×((⌈((1/3))×⌈((5/3)))/(2⌈(((5+1)/3))))=(1/(3a^5 ))×((⌈((1/3))×⌈((2/3)+1))/(1×⌈(1))) =(1/(3a^5 ))×((⌈((1/3))×(2/3)×⌈((2/3)))/1) =(2/(9a^5 ))×⌈((1/3))×⌈(1−(1/3))=(2/(9a^5 ))×(π/(sin((π/3))))=(4/(9a^5 ))×(π/(√3))$
	$1) \int_{0}^{\infty} \frac{d x}{x^{3} + a^{3}} x^{3} = a^{3} {t a n}^{2} α$ $3 x^{2} d x = a^{3} \times 2 t a n α {s e c}^{2} α d α$ $d x = \frac{2 a^{3}}{3} \times \frac{t a n α {s e c}^{2} α}{{(a^{3} {t a n}^{2} α)}^{\frac{2}{3}}} d α = \frac{2 a}{3} \times \frac{{s e c}^{2} α}{{t a n}^{\frac{1}{3}} α} d α$ $\int_{0}^{\frac{π}{2}} \frac{2 a}{3} \times \frac{{s e c}^{2} α}{{t a n}^{\frac{1}{3}} α} \times \frac{1}{a^{3} ({t a n}^{2} α + 1)} \times d α$ $\frac{2}{3 a^{2}} \int_{0}^{\frac{π}{2}} \frac{d α}{{t a n}^{\frac{1}{3}} α}$ $\frac{2}{3 a^{2}} \int_{0}^{\frac{π}{2}} {s i n}^{\frac{- 1}{3}} α {c o s}^{\frac{1}{3}} α d α$ $u s i n g g a m m a b e t a f u n c t i o n . . .$ $\int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} α {c o s}^{2 q - 1} α d α = \frac{⌈ (p) ⌈ q)}{2 (⌈ p + q)}$ $h e r e 2 p - 1 = \frac{- 1}{3} 2 p = \frac{2}{3} p = \frac{1}{3}$ $2 q - 1 = \frac{1}{3} 2 q = \frac{4}{3} q = \frac{2}{3}$ $s o a n s i s \frac{2}{3 a^{2}} \times \frac{⌈ (\frac{1}{3}) \times ⌈ (\frac{2}{3})}{2 ⌈ (\frac{1}{3} + \frac{2}{3})}$ $= \frac{1}{3 a^{2}} \times \frac{⌈ (\frac{1}{3}) \times ⌈ (1 - \frac{1}{3})}{1} = \frac{1}{3 a^{2}} \times \frac{π}{s i n (\frac{π}{3})} = \frac{2 π}{3 \sqrt{3} a^{2}}$ $2) \int_{0}^{\infty} \frac{d x}{{(x^{3} + a^{3})}^{2}} = \int_{0}^{\frac{π}{2}} \frac{2 a}{3} \times \frac{{s e c}^{2} α}{{t a n}^{\frac{1}{3}} α} \times \frac{1}{{a^{3} ({t a n}^{2} α + 1)}^{2}} d α$ $\int_{0}^{\frac{π}{2}} \frac{2 a}{3} \times \frac{{s e c}^{2} α}{{t a n}^{\frac{1}{3}} α} \times \frac{1}{a^{6} \times {s e c}^{4} α} d α$ $\frac{2}{3 a^{5}} \int_{0}^{\frac{π}{2}} \frac{{c o s}^{\frac{1}{3}} α}{{s i n}^{\frac{1}{3}} α} \times {c o s}^{2} α d α$ $\frac{2}{3 a^{5}} \int_{0}^{\frac{π}{2}} {s i n}^{\frac{- 1}{3}} α {c o s}^{\frac{7}{3}} α d α$ $2 p - 1 = \frac{- 1}{3} p = \frac{1}{3}$ $2 q - 1 = \frac{7}{3} 2 q = \frac{10}{3} q = \frac{5}{3}$ $\frac{2}{3 a^{5}} \times \frac{⌈ (\frac{1}{3}) \times ⌈ (\frac{5}{3})}{2 ⌈ (\frac{5 + 1}{3})} = \frac{1}{3 a^{5}} \times \frac{⌈ (\frac{1}{3}) \times ⌈ (\frac{2}{3} + 1)}{1 \times ⌈ (1)}$ $= \frac{1}{3 a^{5}} \times \frac{⌈ (\frac{1}{3}) \times \frac{2}{3} \times ⌈ (\frac{2}{3})}{1}$ $= \frac{2}{9 a^{5}} \times ⌈ (\frac{1}{3}) \times ⌈ (1 - \frac{1}{3}) = \frac{2}{9 a^{5}} \times \frac{π}{s i n (\frac{π}{3})} = \frac{4}{9 a^{5}} \times \frac{π}{\sqrt{3}}$
	Commented bytanmay.chaudhury50@gmail.com last updated on 25/Sep/18

	Commented bytanmay.chaudhury50@gmail.com last updated on 25/Sep/18

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