∫_(−1) ^( 1) (2+x)sin^(−1) (((√(3−3x^2 ))/(2+x)))dx = ?


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Question Number 73689 by ajfour last updated on 14/Nov/19

$\int_{- 1}^{1} (2 + x) \sin^{- 1} (\frac{\sqrt{3 - 3 x^{2}}}{2 + x}) d x = ?$
	Answered by mind is power last updated on 15/Nov/19
	$(d/dx){sin^(−1) (((√(3−3x^2 ))/(2+x)))}=f(x)^′ ={((−3x)/((2+x)(√(3−3x^2 )))) −((√(3−3x^2 ))/((2+x)^2 ))}.(1/(√(1−((3−3x^2 )/((2+x)^2 ))))) ={((−3x(2+x)−(3−3x^2 ))/((2+x)^2 .(√(3−3x^2 ))))}.((2+x)/(√((2x+1)^2 )))={((−6x−3)/(∣2x+1∣))}.(1/((2+x)(√(3−3x^2 )))) =((−3sign(2x+1))/((2+x)(√(3−3x^2 )))) if xε[−(1/2),1] =((−3)/((2+x)(√(3−3x^2 )))),(3/((2+x)(√(3−3x^2 )))) ifx∈[−1,((−1)/2)] ∫_(−1) ^1 (2+x)sin^(−1) (((√(3−3x^2 ))/(2+x)))=[(2x+(x^2 /2))sin^(−1) (((√(3−3x^2 ))/(2+x)))]_(−1) ^1 −∫_(−1) ^(−(1/2)) ((4x+x^2 )/2).(3/((2+x)(√(3−3x^2 ))))−∫_(−(1/2)) ^1 ((4x+x^2 )/2).((−3dx)/((2+x)(√(3−3x^2 )))) all we need is ∫((3(x^2 +4x)dx)/((2+x)(√(3−3x^2 )))),x=sin(θ),θ∈[−(π/2),(π/2)]⇒cos(θ)≥0 ⇒∫((3(sin^2 +4sin(θ))cos(θ)dθ)/({2+sin(θ)}.cos(θ).(√3)))=∫((3(sin(θ)+2)^2 −12)/((√3)(2+sin(θ)}))dθ =(√3)∫(sin(θ)+2)−4(√3)∫(dθ/(2+sin(θ))) ,tg((θ/2))=t in2nd =−(√3)cos(θ)+2(√3)θ−4(√3)∫(2/(2+((2t)/(1+t^2 )))).(dt/(1+t^2 ))=−(√3)cos(θ)+2(√(3 θ)) −4(√3)∫(dt/(t^2 +t+1)) =−(√3)cos(θ)+2(√3)θ−8arctan((((2tg((θ/2)))/(√3))+(1/(√3)))) ∫(((x^2 +4x)dx)/((2+x)(√(3−3x^2 ))))=−(√3) (√(1−x^2 )) +2(√3)arcsin(x)−8arctan(((2tg(((arcsin(x))/2))+1)/(√3))). −{−(√3).((√3)/2)−2(√3)(π/6)−8.(π/(12)).)}+(1/2){−2(√3).(π/2)+8.(π/6)}+(1/2){2(√3).(π/2)−8.(π/3)} =(3/2)+(π/(√3))$
	$\frac{d}{d x} {{s i n}^{- 1} (\frac{\sqrt{3 - 3 x^{2}}}{2 + x})} = f {(x)}^{^{'}} = {\frac{- 3 x}{(2 + x) \sqrt{3 - 3 x^{2}}} - \frac{\sqrt{3 - 3 x^{2}}}{{(2 + x)}^{2}}} . \frac{1}{\sqrt{1 - \frac{3 - 3 x^{2}}{{(2 + x)}^{2}}}}$ $= {\frac{- 3 x (2 + x) - (3 - 3 x^{2})}{{(2 + x)}^{2} . \sqrt{3 - 3 x^{2}}}} . \frac{2 + x}{\sqrt{{(2 x + 1)}^{2}}} = {\frac{- 6 x - 3}{∣ 2 x + 1 ∣}} . \frac{1}{(2 + x) \sqrt{3 - 3 x^{2}}}$ $= \frac{- 3 s i g n (2 x + 1)}{(2 + x) \sqrt{3 - 3 x^{2}}} i f x ϵ [- \frac{1}{2}, 1]$ $= \frac{- 3}{(2 + x) \sqrt{3 - 3 x^{2}}}, \frac{3}{(2 + x) \sqrt{3 - 3 x^{2}}} i f x \in [- 1, \frac{- 1}{2}]$ $\int_{- 1}^{1} (2 + x) {s i n}^{- 1} (\frac{\sqrt{3 - 3 x^{2}}}{2 + x}) = {[(2 x + \frac{x^{2}}{2}) {s i n}^{- 1} (\frac{\sqrt{3 - 3 x^{2}}}{2 + x})]}_{- 1}^{1} - \int_{- 1}^{- \frac{1}{2}} \frac{4 x + x^{2}}{2} . \frac{3}{(2 + x) \sqrt{3 - 3 x^{2}}} - \int_{- \frac{1}{2}}^{1} \frac{4 x + x^{2}}{2} . \frac{- 3 d x}{(2 + x) \sqrt{3 - 3 x^{2}}}$ $a l l w e n e e d i s \int \frac{3 (x^{2} + 4 x) d x}{(2 + x) \sqrt{3 - 3 x^{2}}}, x = s i n (θ), θ \in [- \frac{π}{2}, \frac{π}{2}] \Rightarrow c o s (θ) ⩾ 0$ $\Rightarrow \int \frac{3 ({s i n}^{2} + 4 s i n (θ)) c o s (θ) d θ}{{2 + s i n (θ)} . c o s (θ) . \sqrt{3}} = \int \frac{3 {(s i n (θ) + 2)}^{2} - 12}{\sqrt{3} (2 + s i n (θ)}} d θ$ $= \sqrt{3} \int (s i n (θ) + 2) - 4 \sqrt{3} \int \frac{d θ}{2 + s i n (θ)}, t g (\frac{θ}{2}) = t i n 2 n d$ $= - \sqrt{3} c o s (θ) + 2 \sqrt{3} θ - 4 \sqrt{3} \int \frac{2}{2 + \frac{2 t}{1 + t^{2}}} . \frac{d t}{1 + t^{2}} = - \sqrt{3} c o s (θ) + 2 \sqrt{3 θ} - 4 \sqrt{3} \int \frac{d t}{t^{2} + t + 1}$ $= - \sqrt{3} c o s (θ) + 2 \sqrt{3} θ - 8 a r c t a n ((\frac{2 t g (\frac{θ}{2})}{\sqrt{3}} + \frac{1}{\sqrt{3}}))$ $\int \frac{(x^{2} + 4 x) d x}{(2 + x) \sqrt{3 - 3 x^{2}}} = - \sqrt{3} \sqrt{1 - x^{2}} + 2 \sqrt{3} a r c s i n (x) - 8 a r c t a n (\frac{2 t g (\frac{a r c s i n (x)}{2}) + 1}{\sqrt{3}}) .$ $- {- \sqrt{3} . \frac{\sqrt{3}}{2} - 2 \sqrt{3} \frac{π}{6} - 8 . \frac{π}{12} .)} + \frac{1}{2} {- 2 \sqrt{3} . \frac{π}{2} + 8 . \frac{π}{6}} + \frac{1}{2} {2 \sqrt{3} . \frac{π}{2} - 8 . \frac{π}{3}}$ $= \frac{3}{2} + \frac{π}{\sqrt{3}}$
	Commented by ajfour last updated on 15/Nov/19

	$T h a n k s, P o w e r f u l M i n d!$
	Commented by mind is power last updated on 15/Nov/19

	$y^{'} r e w e l c o m$
	Answered by MJS last updated on 15/Nov/19

	$\int (2 + x) \arcsin \frac{\sqrt{3 - 3 x^{2}}}{2 + x} d x =$ $[t = \arccos \frac{2 x + 1}{x + 2} \to d x = - \frac{(x + 2) \sqrt{3 - 3 x^{2}}}{3} d t]$ $(1)$ $for 0 ⩽ t ⩽ \frac{π}{2}$ $= - 9 \int \frac{t \sin t}{{(2 - \cos t)}^{3}} d t =$ $[\begin{matrix} \int u^{'} v = u v - \int {u v}^{'} \\ u^{'} = \frac{\sin t}{{(2 - \cos t)}^{3}} \Rightarrow u = - \frac{1}{2 {(2 - \cos t)}^{2}} \\ v = t \Rightarrow v^{'} = 1 \end{matrix}]$ $= \frac{9 t}{2 {(2 - \cos t)}^{2}} - \frac{9}{2} \int \frac{d t}{{(2 - \cos t)}^{2}} =$ $\int \frac{d t}{{(2 - \cos t)}^{2}} =$ $[w = \tan \frac{t}{2} \to d t = \frac{2 d w}{w^{2} + 1}]$ $= 2 \int \frac{w^{2} + 1}{{(3 w^{2} + 1)}^{2}} d w =$ $= \frac{2 w}{3 (3 w^{2} + 1)} + \frac{4 \sqrt{3}}{9} \arctan \sqrt{3} w =$ $= \frac{\sin t}{3 (2 - \cos t)} + \frac{4 \sqrt{3}}{9} \arctan (\sqrt{3} \tan \frac{t}{2})$ $= \frac{9 t}{2 {(2 - \cos t)}^{2}} - \frac{3 \sin t}{2 (2 - \cos t)} - 2 \sqrt{3} \arctan (\sqrt{3} \tan \frac{t}{2})$ $(2)$ $for \frac{π}{2} ⩽ t ⩽ π$ $= - 9 π \int \frac{\sin t}{{(2 - \cos t)}^{3}} d t + 9 \int \frac{t \sin t}{{(2 - \cos t)}^{3}} d t =$ $= \frac{9 π}{2 {(2 - \cos t)}^{2}} - (the above)$ $\Rightarrow$ $\underset{- 1}{\int^{1}} (2 + x) \arcsin \frac{\sqrt{3 - 3 x^{2}}}{2 + x} d x =$ $= \frac{\sqrt{3}}{3} π + \frac{3}{2}$
	Commented by ajfour last updated on 15/Nov/19

	$S u p e r b S i r; t h a n k s (y o u a r e$ $i n d e e d a n i n t e g r a l l o v e r)!$
	Commented by ajfour last updated on 15/Nov/19

	$S i r w h a t i f$ $I = \underset{- 1}{\int^{1}} {(2 + x)}^{2} \arcsin \frac{\sqrt{3 - 3 x^{2}}}{2 + x} d x ?$
	Commented by MJS last updated on 15/Nov/19

	$in this case we get$ $- 27 \int \frac{t \sin t}{{(2 - \cos t)}^{4}} d t and (- 27 \int \frac{(π - t) \sin t}{{(2 - \cos t)}^{4}} d t)$ $same path leads to$ $- 27 \int \frac{t \sin t}{{(2 - \cos t)}^{4}} d t =$ $= \frac{9 t}{{(2 - \cos t)}^{3}} + \frac{3 \sin t (5 - 2 \cos t)}{2 {(2 - \cos t)}^{2}} + 3 \sqrt{3} \arctan (\sqrt{3} \tan \frac{t}{2})$ $- 27 \int \frac{(π - t) \sin t}{{(2 - \cos t)}^{4}} d t =$ $= - 27 π \int \frac{\sin t}{{(2 - \cos t)}^{4}} d t + 27 \int \frac{t \sin t}{{(2 - \cos t)}^{4}} =$ $= \frac{9 π}{{(2 - \cos t)}^{3}} - (the above)$ $\Rightarrow$ $\underset{- 1}{\int^{1}} {(2 + x)}^{2} \arcsin \frac{\sqrt{3 - 3 x^{2}}}{2 + x} d x =$ $= \frac{\sqrt{3}}{2} π + \frac{15}{4}$
	Commented by ajfour last updated on 15/Nov/19

	$G r e a t, S i r . A b s o l u t e l y c o r r e c t,$ $I a p p l i e d i t, t h a n k s .$
	Commented by MJS last updated on 15/Nov/19

	${you}^{'} re welcome$
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