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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 by maxmathsup 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 by maxmathsup 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 by maxmathsup by imad last updated on 25/Sep/18

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 by maxmathsup 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) \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 \dots

\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 by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18

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