calculatef-a-a-a-dx-t-2-x-2-3-2-with-a-gt-0-

Question Number 33599 by abdo imad last updated on 19/Apr/18

c a l c u l a t e f (a) = \int_{- a}^{a} \frac{d x}{{(t^{2} + x^{2})}^{\frac{3}{2}}} w i t h a > 0 .

Commented by abdo imad last updated on 20/Apr/18

c a s e 1 t \neq o c h a n g e m e n t x = t t a n θ g i v e

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

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

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

= \frac{2}{t^{2}} \int_{0}^{a r c t a n (\frac{a}{t})} c o s θ d θ = \frac{2}{t^{2}} {[s i n θ]}_{0}^{a r c a n (\frac{a}{t})}

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

f (a) = \frac{2}{t^{2}} \frac{\frac{a}{t}}{\sqrt{1 + \frac{a^{2}}{t^{2}}}} = \frac{2 a}{t} \frac{1}{\sqrt{\frac{t^{2} + a^{2}}{t^{2}}}} = \frac{2 a ∣ t ∣}{t \sqrt{a^{2} + t^{2}}}

f (a) = \frac{2 a ξ (t)}{\sqrt{t^{2} + a^{2}}} w i t h ξ (t) = 1 i f t > 0 a n d ξ (t) = - 1 i f t < 0

c a s e 2 i f t = 0

f (a) = \int_{- a}^{a} \frac{d x}{x^{3}} = 0 b e c a u s e x \to x^{3} i s o d d .

Commented by abdo imad last updated on 20/Apr/18

e r r o r a t l i n e 7 f (a) = \frac{2 a}{t^{2}} \frac{1}{\sqrt{\frac{t^{2} + a^{2}}{t^{2}}}} = \frac{2 a ∣ t ∣}{t^{2} \sqrt{a^{2} + t^{2}}}

= \frac{2 a ξ (t)}{t \sqrt{a^{2} + t^{2}}} .

Answered by alex041103 last updated on 20/Apr/18

\int \frac{d x}{{(t^{2} + x^{2})}^{3 / 2}} = \frac{1}{t^{3}} \int \frac{d x}{{(1 + {(\frac{x}{t})}^{2})}^{3 / 2}}

L e t u = x / t \Rightarrow d x = t d u

\frac{1}{t^{3}} \int \frac{d x}{{(1 + {(\frac{x}{t})}^{2})}^{3 / 2}} = \frac{1}{t^{2}} \int \frac{d u}{{(1 + u^{2})}^{3 / 2}}

W e u s e t h e s t a n d a r t t r i g o n o m e t r i c s u b s t i t u t i o n :

u = t a n θ d u = {s e c}^{2} θ d θ

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

A n d {(1 + {t a n}^{2} θ)}^{3 / 2} = {({s e c}^{2} θ)}^{3 / 2} = {s e c}^{3} θ

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

= \frac{1}{t^{2}} \int c o s θ d θ = \frac{1}{t^{2}} s i n θ

u = t a n θ \Rightarrow θ = a r c t a n (u)

\Rightarrow \frac{1}{t^{2}} s i n θ = \frac{1}{t^{2}} \frac{u}{\sqrt{1 + u^{2}}} = \frac{u}{t \sqrt{t^{2} + {(t u)}^{2}}} =

= \frac{1}{t^{2}} \frac{x}{\sqrt{t^{2} + x^{2}}}

\Rightarrow \int_{- a}^{a} \frac{d x}{{(t^{2} + x^{2})}^{3 / 2}} = \frac{1}{t^{2}} {[\frac{x}{\sqrt{t^{2} + x^{2}}}]}_{- a}^{a} =

= \frac{2 a}{t^{2} \sqrt{t^{2} + a^{2}}}

\Rightarrow \int_{- a}^{a} \frac{d x}{{(t^{2} + x^{2})}^{3 / 2}} = \frac{2 a}{t^{2} \sqrt{t^{2} + a^{2}}}

A n y q u e s t i o n s ?

Commented by abdo imad last updated on 20/Apr/18

y o u h a v e c o m m i t e d a e r r o r s i r a l e x .

Commented by alex041103 last updated on 20/May/18

C a n y o u p o i n t i t o u t, s o I c a n f i x i t .