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Question Number 56383 by Tawa1 last updated on 15/Mar/19

Commented by Tawa1 last updated on 16/Mar/19

$God bless you sir$
Commented by maxmathsup by imad last updated on 15/Mar/19
$∫_(−1) ^7 (dx/((^3 (√(x+1)))))dx =_(x+1 =t^3 ) ∫_0 ^(2(√2)) ((3t^2 dt)/t) =3 ∫_0 ^(2(√2)) tdt =3[(t^2 /2)]_0 ^(2(√2)) =(3/2){2(√2))^2 =12.$
$\int_{- 1}^{7} \frac{d x}{(^{3} \sqrt{x + 1})} d x =_{x + 1 = t^{3}} \int_{0}^{2 \sqrt{2}} \frac{3 t^{2} d t}{t} = 3 \int_{0}^{2 \sqrt{2}} t d t = 3 {[\frac{t^{2}}{2}]}_{0}^{2 \sqrt{2}}$ $= \frac{3}{2} {{2 \sqrt{2})}^{2} = 12 .$
Commented by maxmathsup by imad last updated on 16/Mar/19

$w e c a n u s e t h e r e s u l t \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} w i t h 0 < a < 1 s o$ $c h a n g e m e n t x = t^{\frac{1}{4}} g i v e \int_{0}^{\infty} \frac{d x}{1 + x^{4}} = \int_{0}^{\infty} \frac{1}{4 (1 + t)} t^{\frac{1}{4} - 1} d t$ $= \frac{1}{4} \frac{π}{s i n (\frac{π}{4})} = \frac{π}{4 \frac{\sqrt{2}}{2}} = \frac{π}{2 \sqrt{2}} .$
Commented by Abdo msup. last updated on 16/Mar/19

$t h a n k y o u s o m u c h s i r .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Mar/19

	$\int \frac{d x}{x^{4} + 1}$ $\frac{1}{2} \int \frac{\frac{2}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x$ $\frac{1}{2} \int \frac{1 + \frac{1}{x^{2}} - (1 - \frac{1}{x^{2}})}{x^{2} + \frac{1}{x^{2}}} d x$ $\frac{1}{2} \int \frac{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + 2} - \frac{1}{2} \int \frac{d (x + \frac{1}{x})}{{(x + \frac{1}{x})}^{2} - 2}$ $∣ \frac{1}{2} \times \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}}) - \frac{1}{2} \times \frac{1}{2 \sqrt{2}} l n (\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}}) ∣_{0}^{\infty}$ $∣ \frac{1}{2 \sqrt{2}} {t a n}^{- 1} (\frac{1 - \frac{1}{x^{2}}}{\frac{\sqrt{2}}{x}}) - \frac{1}{4 \sqrt{2}} l n (\frac{1 + \frac{1}{x^{2}} - \frac{\sqrt{2}}{x}}{1 + \frac{1}{x^{2}} + \frac{\sqrt{2}}{x}}) ∣_{0}^{\infty}$ $n o w p u t t i n g x = \infty \to \frac{1}{2 \sqrt{2}} {t a n}^{- 1} (\frac{1 - 0}{0}) - \frac{1}{4 \sqrt{2}} l n (\frac{1 + 0 - 0}{1 + 0 + 0})$ $= \frac{1}{2 \sqrt{2}} \times \frac{π}{2} = \frac{π}{4 \sqrt{2}}$ $n o w a g a i n a d j u s t m e n t t o p u t x = 0$ $\frac{1}{2 \sqrt{2}} {t a n}^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}}) - \frac{1}{4 \sqrt{2}} l n (\frac{x^{2} + 1 - x \sqrt{2}}{x^{2} + 1 + x \sqrt{2}}) n o w p u t$ $x = 0$ $\frac{1}{2 \sqrt{2}} {t a n}^{- 1} (- \infty) \to \frac{1}{2 \sqrt{2}} \times (- \frac{π}{2})$ $s o a n s w e r i s$ $\frac{π}{4 \sqrt{2}} - (- \frac{π}{4 \sqrt{2}}) \to 2 \times \frac{π}{4 \sqrt{2}} = \frac{π}{2 \sqrt{2}}$
	Commented by Tawa1 last updated on 15/Mar/19

	$God bless you sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Mar/19

	$\lim_{a \to - 1} \int_{a}^{7} \frac{d x}{{(1 + x)}^{\frac{1}{3}}}$ $\lim_{a \to - 1} ∣ \frac{{(1 + x)}^{\frac{- 1}{3} + 1}}{\frac{2}{3}} ∣_{a}^{7}$ $= \lim_{a \to - 1} [\frac{3}{2} {(8)}^{\frac{2}{3}} - \frac{3}{2} {(1 + a)}^{\frac{2}{3}}]$ $= \frac{3}{2} \times 4 \to 6 a n s$
	Commented by Tawa1 last updated on 15/Mar/19

	$God bless you sir .$
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