find the value of ∫_0 ^(√3) arcsin(((2t)/(1+t^2 )))dt


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Question Number 41702 by abdo.msup.com last updated on 11/Aug/18

$f i n d t h e v a l u e o f \int_{0}^{\sqrt{3}} a r c s i n (\frac{2 t}{1 + t^{2}}) d t$
Commented by math khazana by abdo last updated on 12/Aug/18
$let A = ∫_0 ^(√3) arcsin(((2t)/(1+t^2 )))dt changement t =tanθ give A = ∫_0 ^(π/3) arcsin(sin(2θ))(1+tan^2 θ)dθ =_(2θ=u) ∫_0 ^((2π)/3) arcsin(sinu)(1+tan^2 ((u/2)))du =∫_0 ^(π/2) u(1+tan^2 ((u/2)))du + ∫_(π/2) ^((2π)/3) arcsin(sinu)(1+tan^2 ((u/2)))du ∫_0 ^(π/2) u(1+tan^2 ((u/2)))du =[(u^2 /2)]_0 ^(π/2) +∫_0 ^(π/2) tan^2 ((u/2))du =(π^2 /8) + ∫_0 ^(π/2) (1+tan^2 ((u/2)))du −(π/2) = [2tan((u/2))]_0 ^(π/2) +(π^2 /8) −(π/2) =2 +(π^2 /8) −(π/2) also ∫_(π/2) ^((2π)/3) arcsin(sinu)(1+tan^2 ((u/2)))du =_(u=π−t) −∫_(π/2) ^(π/3) arcsin(sin(π−t))(1+tan^2 ((π/2)−(t/2)))dt = ∫_(π/3) ^(π/2) t{1 +(1/(tan^2 ((t/2))))}dt ( bijection at [−(π/2),(π/2)] = [(t^2 /2)]_(π/3) ^(π/2) + ∫_(π/3) ^(π/2) (t/(tan^2 ((t/2))))dt =( (π^2 /8) −(π^2 /(18))) + ∫_(π/3) ^(π/2) (t/(tan^2 ((t/2))))dt but ∫_(π/3) ^(π/2) (t/(tan^2 ((t/2))))dt =_((t/2)=u) ∫_(π/6) ^(π/4) ((2u)/(tan^2 (u))) 2du =4 ∫_(π/6) ^(π/4) (u/(tan^2 u)) du =_(tanu =α) 4 ∫_(1/(√3)) ^1 ((arctan(α))/α^2 ) (dα/(1+α^2 )) =4 ∫_(1/(√3)) ^1 ((arctan(α))/(α^2 (1+α^2 ))) dα ....be continued...$
$l e t A = \int_{0}^{\sqrt{3}} a r c s i n (\frac{2 t}{1 + t^{2}}) d t c h a n g e m e n t$ $t = t a n θ g i v e$ $A = \int_{0}^{\frac{π}{3}} a r c s i n (s i n (2 θ)) (1 + {t a n}^{2} θ) d θ$ $=_{2 θ = u} \int_{0}^{\frac{2 π}{3}} a r c s i n (s i n u) (1 + {t a n}^{2} (\frac{u}{2})) d u$ $= \int_{0}^{\frac{π}{2}} u (1 + {t a n}^{2} (\frac{u}{2})) d u + \int_{\frac{π}{2}}^{\frac{2 π}{3}} a r c s i n (s i n u) (1 + {t a n}^{2} (\frac{u}{2})) d u$ $\int_{0}^{\frac{π}{2}} u (1 + {t a n}^{2} (\frac{u}{2})) d u = {[\frac{u^{2}}{2}]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} {t a n}^{2} (\frac{u}{2}) d u$ $= \frac{π^{2}}{8} + \int_{0}^{\frac{π}{2}} (1 + {t a n}^{2} (\frac{u}{2})) d u - \frac{π}{2}$ $= {[2 t a n (\frac{u}{2})]}_{0}^{\frac{π}{2}} + \frac{π^{2}}{8} - \frac{π}{2} = 2 + \frac{π^{2}}{8} - \frac{π}{2} a l s o$ $\int_{\frac{π}{2}}^{\frac{2 π}{3}} a r c s i n (s i n u) (1 + {t a n}^{2} (\frac{u}{2})) d u$ $=_{u = π - t} - \int_{\frac{π}{2}}^{\frac{π}{3}} a r c s i n (s i n (π - t)) (1 + {t a n}^{2} (\frac{π}{2} - \frac{t}{2})) d t$ $= \int_{\frac{π}{3}}^{\frac{π}{2}} t {1 + \frac{1}{{t a n}^{2} (\frac{t}{2})}} d t (b i j e c t i o n a t [- \frac{π}{2}, \frac{π}{2}]$ $= {[\frac{t^{2}}{2}]}_{\frac{π}{3}}^{\frac{π}{2}} + \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{t}{{t a n}^{2} (\frac{t}{2})} d t$ $= (\frac{π^{2}}{8} - \frac{π^{2}}{18}) + \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{t}{{t a n}^{2} (\frac{t}{2})} d t b u t$ $\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{t}{{t a n}^{2} (\frac{t}{2})} d t =_{\frac{t}{2} = u} \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{2 u}{{t a n}^{2} (u)} 2 d u$ $= 4 \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{u}{{t a n}^{2} u} d u =_{t a n u = α} 4 \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n (α)}{α^{2}} \frac{d α}{1 + α^{2}}$ $= 4 \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n (α)}{α^{2} (1 + α^{2})} d α . . . . b e c o n t i n u e d . . .$
Commented by alex041103 last updated on 12/Aug/18

$Y o u a r e w r o n g . W h e n y o u m a k e t h e$ $s u b s t i t u t i o n t = t a n θ, y o u h a v e t o m u l t i p l y$ $t h e i n t e g r a n d b y a f a c t o r o f {s e c}^{2} θ,$ $b e c a u s e d t = {s e c}^{2} θ d θ .$
Commented by math khazana by abdo last updated on 12/Aug/18

$y e s y e s y o u a r e r i g h t t h a n k s . . .$
	Answered by alex041103 last updated on 11/Aug/18

	$l e t t = t a n (x / 2) d t = \frac{{s e c}^{2} (x / 2)}{2} d x \Rightarrow x = 2 a r c t a n (t)$ $\frac{2 t}{1 + t^{2}} = s i n (x) \Rightarrow \int_{0}^{\frac{2 π}{3}} x d (t a n (x / 2))$ $I B P$ $\int_{0}^{\frac{2 π}{3}} x d (t a n (x / 2)) = {[x t a n (x / 2)]}_{0}^{2 π / 3} - 2 \int_{0}^{2 π / 3} t a n (x / 2) d (x / 2) =$ $= 2 \sqrt{3} π - 2 \int_{0}^{π / 3} t a n (u) d u$ $B u t \int t a n (x) d x = l n (s e c x) + C$ $\Rightarrow \int_{0}^{\sqrt{3}} a r c s i n (\frac{2 t}{1 + t^{2}}) d t = 2 \sqrt{3} π - 2 l n (\frac{s e c (π / 3)}{s e c (0)}) =$ $= 2 \sqrt{3} π - 2 l n (2)$ $\int_{0}^{\sqrt{3}} a r c s i n (\frac{2 t}{1 + t^{2}}) d t = 2 \sqrt{3} π - 2 l n (2)$
	Commented by turbo msup by abdo last updated on 12/Aug/18

	$s i r a l e x y o u m u s t d i v i d e t h e$ $i n t e g r a l b e c a u s e a r c s i n x i s$ $d e f i n e d f r o m [- 1, 1] t o [- \frac{π}{2}, \frac{π}{2}] (b i j e c t i o n)$ $s o y o u r f i n a l r e s u l t i s n o t c o r r e c t$
	Commented by alex041103 last updated on 12/Aug/18

	$I n t h a t c a s e i t i s p e r f e c t l y O K .$ $B e c a u s e i n t h e R$ $a r c s i n (\frac{2 t}{1 + t^{2}}) \in [- 1, 1]$
	Commented by alex041103 last updated on 12/Aug/18

	Commented by alex041103 last updated on 12/Aug/18

	$O h . . . I u n d e r s t a n d w h a t y o u m e a n .$ ${Y o u}^{'} r e a c t u a l l y r i g h t . B u t . . .$ $t h e r e a r e d i f f e r e n t a r c s i n f u n c t i o n s .$ $f o r e x a m p l e t h e r e i s o n e f o r w h i c h$ $a r c s i n (x) \in [0, 2 π) . . . I t h o u g h t t h a t o n e$ $w a s t h e o n e i n t h e p r o b l e m . . . . a m i s s u n d e r s t a n d i n g . .$ $t h a n k y o u f o r p o i n t i n g i t o u t f o r m e . . .$
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