find f(a) = ∫ (dx/(√(1−ax^2 ))) with a from R .


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Question Number 36818 by maxmathsup by imad last updated on 06/Jun/18

$f i n d f (a) = \int \frac{d x}{\sqrt{1 - {a x}^{2}}} w i t h a f r o m R .$
Commented by prof Abdo imad last updated on 06/Jun/18
$case 1 a>0 changement (√a)x= sint give f(a) = ∫ (1/(cost)) ((cost dt)/(√a)) =(1/(√a)) arcsin((√a) x) +c case 2 a<0 changement (√(−a))x=sh(t) give f(a) = ∫ (1/(ch(t))) ((ch(t))/(√(−a))) = (1/(√(−a))) argsh((√(−a))x) =(1/(√(−a))) ln{ x(√(−a)) +(√(1−ax^2 )) ) +c .$
$c a s e 1 a > 0 c h a n g e m e n t \sqrt{a} x = s i n t g i v e$ $f (a) = \int \frac{1}{c o s t} \frac{c o s t d t}{\sqrt{a}} = \frac{1}{\sqrt{a}} a r c s i n (\sqrt{a} x) + c$ $c a s e 2 a < 0 c h a n g e m e n t \sqrt{- a} x = s h (t) g i v e$ $f (a) = \int \frac{1}{c h (t)} \frac{c h (t)}{\sqrt{- a}} = \frac{1}{\sqrt{- a}} a r g s h (\sqrt{- a} x)$ $= \frac{1}{\sqrt{- a}} l n {x \sqrt{- a} + \sqrt{1 - {a x}^{2}}) + c .$
	Answered by MJS last updated on 06/Jun/18
	$∫(dx/(√(1−ax^2 )))= [t=(√a)x → dx=(dt/(√a))] =(1/(√a))∫(dt/(√(1−t^2 )))=(1/(√a))arcsin t= ((√a)/a)arcsin (√a)x +C { ((a>0 (1/(√a))arcsin (√a)x +C)),((a=0 x+C)),((a<0 (1/(√(−a)))arcsinh (√(−a))x +C)) :}$
	$\int \frac{d x}{\sqrt{1 - {a x}^{2}}} =$ $[t = \sqrt{a} x \to d x = \frac{d t}{\sqrt{a}}]$ $= \frac{1}{\sqrt{a}} \int \frac{d t}{\sqrt{1 - t^{2}}} = \frac{1}{\sqrt{a}} \arcsin t =$ $\frac{\sqrt{a}}{a} \arcsin \sqrt{a} x + C$ ${\begin{cases} a > 0 \frac{1}{\sqrt{a}} \arcsin \sqrt{a} x + C \\ a = 0 x + C \\ a < 0 \frac{1}{\sqrt{- a}} arcsinh \sqrt{- a} x + C \end{cases}$
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