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Question Number 54011 by maxmathsup by imad last updated on 27/Jan/19

c a l c u l a t e f (a) = \int \frac{d x}{\sqrt{1 + a x} - \sqrt{1 - a x}} w i t h a > 0 .

2) c a l c u l a t e U_{n} = \int_{- \frac{1}{n a}}^{\frac{1}{n a}} \frac{d x}{\sqrt{1 + a x} - \sqrt{1 - a x}} w i t h n f r o m N a n d n > 1

f i n d {l i m}_{n \to + \infty} U_{n} a n d s t u d y t h e c o n v e r g e n c e o f Σ U_{n}

Commented by maxmathsup by imad last updated on 28/Jan/19

1) w e h a v e f (a) =_{a x = t} \int \frac{d t}{a (\sqrt{1 + t} - \sqrt{1 - t})} \Rightarrow

a f (a) = \int \frac{\sqrt{1 + t} + \sqrt{1 - t}}{2 t} d t = \frac{1}{2} \int \frac{\sqrt{1 + t}}{t} d t + \frac{1}{2} \int \frac{\sqrt{1 - t}}{t} d t \Rightarrow

2 a f (a) = \int \frac{\sqrt{1 + t}}{t} d t + \int \frac{\sqrt{1 - t}}{t} d t b u t

\int \frac{\sqrt{1 + t}}{t} d t =_{1 + t = u^{2}} \int \frac{u}{u^{2} - 1} (2 u) d u = 2 \int \frac{u^{2}}{u^{2} - 1} d u

= 2 \int \frac{u^{2} - 1 + 1}{u^{2} - 1} = 2 u + 2 \int \frac{d u}{u^{2} - 1} = 2 u + \int (\frac{1}{1 - u} + \frac{1}{1 + u}) d u

= 2 u + l n ∣ \frac{1 + u}{1 - u} ∣ + c_{0} = 2 \sqrt{1 + t} + l n ∣ \frac{1 + \sqrt{1 + t}}{1 - \sqrt{1 - t}} ∣ + c_{0}

= 2 \sqrt{1 + a x} + l n ∣ \frac{1 + \sqrt{1 + a x}}{1 - \sqrt{1 - a x}} ∣ + c_{0} a l s o w e h a v e

\int \frac{\sqrt{1 - t}}{t} d t =_{1 - t = u^{2}} \int \frac{u}{1 - u^{2}} (- 2 u) d u = 2 \int \frac{u^{2}}{u^{2} - 1} \Rightarrow

2 a f (a) = 2 \int \frac{\sqrt{1 + t}}{t} d t \Rightarrow f (a) = \frac{1}{a} {2 \sqrt{1 + a x} + l n ∣ \frac{1 + \sqrt{1 + a x}}{1 - \sqrt{1 - a x}} ∣} .

f (a) = \frac{1}{a} {2 \sqrt{1 + a x} + l n ∣ \frac{1 + \sqrt{1 + a x}}{1 - \sqrt{1 - a x}} ∣} + C .

Commented by maxmathsup by imad last updated on 28/Jan/19

2) w e h a v e U_{n} = \int_{- \frac{1}{n a}}^{\frac{1}{n a}} \frac{d x}{\sqrt{1 + a x} - \sqrt{1 - a x}} = 2 {l i m}_{ξ \to 0} \int_{ξ}^{\frac{1}{n a}} \frac{d x}{\sqrt{1 + a x} - \sqrt{1 - a x}}

b u t \int_{ξ}^{\frac{1}{n a}} \frac{d x}{\sqrt{1 + a x} - \sqrt{1 - a x}} = {[\frac{1}{a} (2 \sqrt{1 + a x} + l n ∣ \frac{1 + \sqrt{1 + a x}}{1 - \sqrt{1 - a x}} ∣)]}_{x = ξ}^{\frac{1}{n a}}

= \frac{1}{a} {2 \sqrt{1 + \frac{1}{n}} + l n ∣ \frac{1 + \sqrt{1 + \frac{1}{n}}}{1 - \sqrt{1 - \frac{1}{n}}} ∣ - 2 \sqrt{1 + a ξ} - l n ∣ \frac{1 + \sqrt{1 + a ξ}}{1 - \sqrt{1 - a ξ}} ∣}

b u t w e c a n t f i n d l i m U_{n} f r o m t h i s q u a n t i t y b e c o n t i n u e d \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19

\int \frac{\sqrt{1 + a x} + \sqrt{1 - a x}}{2 a x} d x

\int \frac{\sqrt{1 + a x}}{2 a x} d x + \int \frac{\sqrt{1 - a x}}{2 a x} d x

t_{1}^{2} = 1 + a x 2 t_{1} d t = a d x

t_{2}^{2} = 1 - a x 2 t_{2} d t = - a d x

\int \frac{t_{1} \times 2 t_{1} d t}{2 a (t_{1}^{2} - 1)} + \int \frac{t_{2} \times - 2 t_{2}}{2 a \times (1 - t_{1}^{2})} {d t}_{2}

\frac{1}{a} \int \frac{t_{1}^{2} - 1 + 1}{t_{1}^{2} - 1} {d t}_{1} + \frac{1}{a} \int \frac{1 - t_{2}^{2} - 1}{1 - t_{2}^{2}} {d t}_{2}

\frac{1}{a} [\int {d t}_{1} + \int \frac{{d t}_{1}}{t_{2}^{2} - 1} + \int {d t}_{2} - \int \frac{{d t}_{2}}{1 - t_{2}^{2}}]

n o w u s e f o r m u l a \dots