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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 by prof Abdo imad last updated on 20/Jun/19

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 by prof Abdo imad last updated on 20/Jun/19

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 by prof 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 by prof Abdo imad last updated on 20/Jun/19

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

= \dots .

= \frac{1}{2 ξ} [\frac{1}{\sqrt{ξ}} \sin^{- 1} (\sqrt{ξ} x) - x \sqrt{1 - ξ x^{2}}] + C

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