find-the-value-of-f-a-a-dx-x-2-3-2-R-

Question Number 35832 by abdo mathsup 649 cc last updated on 24/May/18

f i n d t h e v a l u e o f f (λ) = \int_{- a}^{a} \frac{d x}{{(λ +_{} x^{2})}^{\frac{3}{2}}}

λ \in R .

Commented by abdo.msup.com last updated on 25/May/18

c a s e 1 i f λ > 0 l e t u s e t h e c h a n g e m e n t

x = \sqrt{λ} t a n (t)

f (λ) = \int_{- a r c t a n (\frac{a}{λ})}^{a r c t a n (\frac{a}{λ})} \frac{\sqrt{λ}}{λ^{\frac{3}{2}} {(1 + {t a n}^{2} (t))}^{\frac{3}{2}}} (1 + {t a n}^{2} t) d t

= \frac{1}{λ} \int_{- a r c t a n (\frac{a}{\sqrt{λ}})}^{a r c t a n (\frac{a}{\sqrt{λ}})} \frac{1}{{(1 + {t a n}^{2} t)}^{\frac{1}{2}}} d t

= \frac{2}{λ} \int_{0}^{a r c t a n (\frac{a}{\sqrt{λ}})} c o s (t) d t

= \frac{2}{λ} {[s i n (t)]}_{0}^{a r c t a n (\frac{a}{\sqrt{λ}})}

= \frac{2}{λ} s i n (a r c t a n (\frac{a}{\sqrt{λ}}))

b u t w e h a v e s i n (a r c t a n x) = \frac{x}{\sqrt{1 + x^{2}}}

s o s i n (a r c t a n (\frac{a}{\sqrt{λ}})) = \frac{a}{\sqrt{λ} \sqrt{1 + \frac{a^{2}}{λ}}}

= \frac{a}{\sqrt{λ + a^{2}}} \Rightarrow f (λ) = \frac{2 a}{λ \sqrt{λ + a^{2}}}

Commented by prof Abdo imad last updated on 25/May/18

c a s e 2 i f λ < 0 \Rightarrow - λ > 0 l e t p u t α = - λ

f (λ) = \int_{- a}^{a} \frac{d x}{{(x^{2} - α)}^{\frac{3}{2}}} . c h a n g e m e n t x = \sqrt{α} c h (t)

g i v e

f (λ) = \int_{a r g c h (\frac{- a}{\sqrt{α}})}^{a r g c h (\frac{a}{\sqrt{α}})} \frac{\sqrt{α} s h (t)}{α^{\frac{3}{2}} {({c h}^{2} (t) - 1)}^{\frac{3}{2}}} d t

= \frac{1}{α} \int_{a r g c h (\frac{- a}{\sqrt{α}})}^{a r g c h (\frac{a}{\sqrt{α}})} \frac{s h (t)}{{s h}^{3} (t)} d t

= \frac{2}{α} \int_{a r g c h (\frac{- a}{\sqrt{α}})}^{a r g c h (\frac{a}{\sqrt{α}})} \frac{d t}{c h (2 t) - 1}

= \frac{2}{α} \int_{l n (- \frac{a}{\sqrt{κ}} + \sqrt{\frac{a^{2}}{α} - 1})}^{l n (\frac{a}{\sqrt{α}} + \sqrt{\frac{a^{2}}{α} - 1}} \frac{d t}{\frac{e^{2 t} + e^{- 2 t}}{2} - 1}

= \frac{4}{α} \int_{l n (- \frac{a}{\sqrt{α}} + \sqrt{\frac{a^{2}}{α} - 1})}^{l n (\frac{a}{\sqrt{α}} + \sqrt{\frac{a^{2}}{α} - 1})} \frac{d t}{e^{2 t} + e^{- 2 t} - 2} \dots

b e c o n t i n u e d \dots

Commented by prof Abdo imad last updated on 26/May/18

l e t f i n d K = \int \frac{d t}{e^{2 t} + e^{- 2 t} - 2} c h a m g e m e n t

e^{2 t} = u g i v e K = \int \frac{1}{u + u^{- 1} - 2} \frac{d u}{2 u}

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

= - \frac{1}{2 (u - 1)} = \frac{- 1}{2 (e^{2 t} - 1)} + c s o

f (λ) = - \frac{4}{2 α} {[\frac{1}{e^{2 t} - 1}]}_{l n (\frac{- a}{\sqrt{α}} + \sqrt{\frac{a^{2}}{α} - 1}}^{l n (\frac{a}{\sqrt{α}} + \sqrt{\frac{a^{2}}{α} - 1}}

= \frac{- 2}{α} {\frac{1}{(\frac{a}{\sqrt{α}} + \sqrt{\frac{a^{2}}{α} - 1})} - \frac{1}{(- \frac{a}{\sqrt{α}} + \sqrt{\frac{a^{2}}{α} - 1})}

= \frac{2}{λ} {\frac{1}{(\frac{a}{\sqrt{- λ}} + \sqrt{- \frac{a^{2}}{λ} - 1})} - \frac{1}{(\frac{- a}{\sqrt{- λ}} + \sqrt{- \frac{a^{2}}{λ} - 1})}

w i t h λ < 0 .

Answered by tanmay.chaudhury50@gmail.com last updated on 24/May/18

x = \sqrt{λ} t a n θ d x = \sqrt{λ} {s e c}^{2} θ d θ

I = \int \frac{\sqrt{λ} {s e c}^{2} θ d θ}{{λ (1 + {t a n}^{2} θ)}^{\frac{3}{2}}}

\int \frac{1}{λ} \times \frac{{s e c}^{2} θ}{{s e c}^{3} θ} \times \frac{d θ}{}

= \frac{1}{λ} \int c o s θ d θ

= \frac{1}{λ} s i n θ n o w p u t t i n g t h e l i m i t

= \frac{1}{λ} \times ∣ \frac{x}{{(λ + x^{2})}^{\frac{1}{2}}} ∣_{- a}^{a}

= \frac{1}{λ} \times \frac{2 a}{{(λ + a^{2})}^{\frac{1}{2}}}

Commented by abdo.msup.com last updated on 25/May/18

y o u r a n s w e r i s c o r r e c t s i r T a n m a y

b u t y o u m u s t s t u d y t h e c a s e λ < 0 \dots .