let f(ξ) =∫ (x^2 /(√(1−ξx^2 )))dx with 0<ξ<1 1) determine a explicit form of f(ξ) 2) calculate lim_(ξ→1) f(ξ) 3) calculate ∫_0 ^(1/2) (x^2 /(√(1−sin^2 θ x^2 ))) dx with 0<θ<(π/2)


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Question Number 62342 by maxmathsup by imad last updated on 20/Jun/19

$l e t f (ξ) = \int \frac{x^{2}}{\sqrt{1 - ξ x^{2}}} d x w i t h 0 < ξ < 1$ $1) d e t e r m i n e a e x p l i c i t f o r m o f f (ξ)$ $2) c a l c u l a t e {l i m}_{ξ \to 1} f (ξ)$ $3) c a l c u l a t e \int_{0}^{\frac{1}{2}} \frac{x^{2}}{\sqrt{1 - {s i n}^{2} θ x^{2}}} d x w i t h 0 < θ < \frac{π}{2}$
Commented byprof Abdo imad last updated on 20/Jun/19
$1) f(ξ)=(1/ξ) ∫ ((((√ξ)x)^2 )/(√(1−((√ξ)x)^2 ))) dx =_((√ξ)x=t) (1/ξ) ∫ (t^2 /(√(1−t^2 ))) (dt/(√ξ)) =(1/(ξ(√ξ))) ∫ (t^2 /(√(1−t^2 ))) dt ∫ (t^2 /(√(1−t^2 ))) dt =−∫ ((1−t^2 −1)/(√(1−t^2 )))dt =−∫ (√(1−t^2 ))dt +∫ (dt/(√(1−t^2 ))) +c ∫ (dt/(√(1−t^2 ))) dt =arcsint ∫ (√(1−t^2 ))dt =_(t=sinu) ∫ cos^2 udu =∫ ((1+cos(2u))/2) du =(u/2) +(1/4)sin(2u) =(u/2) +(1/2)sinu cosu =(u/2) +(1/2)sinu(√(1−sin^2 u)) =((arcsint)/2) +(t/2)(√(1−t^2 )) ⇒ f(ξ) =(1/(ξ(√ξ)))(−((arcsint)/2) +(1/2)t(√(1−t^2 )) +arcsint} +c f(ξ)=(1/(ξ(√ξ))){((arcsint)/2) +((t(√(1−t^2 )))/2)} +c$
$1) f (ξ) = \frac{1}{ξ} \int \frac{{(\sqrt{ξ} x)}^{2}}{\sqrt{1 - {(\sqrt{ξ} x)}^{2}}} d x$ $=_{\sqrt{ξ} x = t} \frac{1}{ξ} \int \frac{t^{2}}{\sqrt{1 - t^{2}}} \frac{d t}{\sqrt{ξ}} = \frac{1}{ξ \sqrt{ξ}} \int \frac{t^{2}}{\sqrt{1 - t^{2}}} d t$ $\int \frac{t^{2}}{\sqrt{1 - t^{2}}} d t = - \int \frac{1 - t^{2} - 1}{\sqrt{1 - t^{2}}} d t$ $= - \int \sqrt{1 - t^{2}} d t + \int \frac{d t}{\sqrt{1 - t^{2}}} + c$ $\int \frac{d t}{\sqrt{1 - t^{2}}} d t = a r c s i n t$ $\int \sqrt{1 - t^{2}} d t =_{t = s i n u} \int {c o s}^{2} u d u$ $= \int \frac{1 + c o s (2 u)}{2} d u = \frac{u}{2} + \frac{1}{4} s i n (2 u)$ $= \frac{u}{2} + \frac{1}{2} s i n u c o s u = \frac{u}{2} + \frac{1}{2} s i n u \sqrt{1 - {s i n}^{2} u}$ $= \frac{a r c s i n t}{2} + \frac{t}{2} \sqrt{1 - t^{2}} \Rightarrow$ $f (ξ) = \frac{1}{ξ \sqrt{ξ}} (- \frac{a r c s i n t}{2} + \frac{1}{2} t \sqrt{1 - t^{2}} + a r c s i n t} + c$ $f (ξ) = \frac{1}{ξ \sqrt{ξ}} {\frac{a r c s i n t}{2} + \frac{t \sqrt{1 - t^{2}}}{2}} + c$
Commented byprof Abdo imad last updated on 20/Jun/19
$t =x(√ξ) ⇒ f(ξ) =(1/(2ξ(√ξ))){ arcsin(x(√ξ))+x(√ξ)(√(1−ξx^2 ))} +c$
$t = x \sqrt{ξ} \Rightarrow$ $f (ξ) = \frac{1}{2 ξ \sqrt{ξ}} {a r c s i n (x \sqrt{ξ}) + x \sqrt{ξ} \sqrt{1 - ξ x^{2}}} + c$
Commented byprof Abdo imad last updated on 20/Jun/19

$2) {l i m}_{ξ \to 1} f (ξ) = \frac{1}{2} (a r c s i n (x) + x \sqrt{1 - x^{2}}) + c$
Commented byprof Abdo imad last updated on 20/Jun/19
$3) ∫_0 ^(1/2) (x^2 /(√(1−sin^2 θ x^2 )))dx (ξ=sin^2 θ) =[(1/(2sin^3 θ)){arcsin(xsinθ) +xsinθ(√(1−sin^2 θ x^2 ))}]_0 ^(1/2) =(1/(2sin^3 θ)){ arcsin(((sinθ)/2))+(1/2)sinθ(√(1−(1/4)sin^2 θ))}$
$3) \int_{0}^{\frac{1}{2}} \frac{x^{2}}{\sqrt{1 - {s i n}^{2} θ x^{2}}} d x (ξ = {s i n}^{2} θ)$ $= {[\frac{1}{2 {s i n}^{3} θ} {a r c s i n (x s i n θ) + x s i n θ \sqrt{1 - {s i n}^{2} θ x^{2}}}]}_{0}^{\frac{1}{2}}$ $= \frac{1}{2 {s i n}^{3} θ} {a r c s i n (\frac{s i n θ}{2}) + \frac{1}{2} s i n θ \sqrt{1 - \frac{1}{4} {s i n}^{2} θ}}$
	Answered by mr W last updated on 20/Jun/19

	$f (ξ) = \int \frac{x^{2}}{\sqrt{1 - ξ x^{2}}} d x$ $= \frac{1}{ξ} \int \frac{{(\sqrt{ξ} x)}^{2}}{\sqrt{1 - {(\sqrt{ξ} x)}^{2}}} d (\sqrt{ξ} x)$ $= \frac{1}{ξ} \int \frac{t^{2}}{\sqrt{1 - t^{2}}} d t$ $= \frac{1}{ξ} \int [\frac{1}{\sqrt{1 - t^{2}}} - \sqrt{1 - t^{2}}] d t$ $= . . . .$ $= \frac{1}{2 ξ} [\frac{1}{\sqrt{ξ}} \sin^{- 1} (\sqrt{ξ} x) - x \sqrt{1 - ξ x^{2}}] + C$
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