calculatef(a)= ∫_(−a) ^a (dx/((t^2 +x^2 )^(3/2) )) with a>0 .


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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

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