Question Number 73689 by ajfour last updated on 14/Nov/19
$$\int_{−\mathrm{1}} ^{\:\:\mathrm{1}} \left(\mathrm{2}+{x}\right)\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\right){dx}\:=\:? \\ $$
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))](Q73698.png)
$$\frac{{d}}{{dx}}\left\{{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\right)\right\}={f}\left({x}\right)^{'} =\left\{\frac{−\mathrm{3}{x}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}\:−\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\left(\mathrm{2}+{x}\right)^{\mathrm{2}} }\right\}.\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\frac{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }{\left(\mathrm{2}+{x}\right)^{\mathrm{2}} }}} \\ $$$$=\left\{\frac{−\mathrm{3}{x}\left(\mathrm{2}+{x}\right)−\left(\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} \right)}{\left(\mathrm{2}+{x}\right)^{\mathrm{2}} .\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}\right\}.\frac{\mathrm{2}+{x}}{\sqrt{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }}=\left\{\frac{−\mathrm{6}{x}−\mathrm{3}}{\mid\mathrm{2}{x}+\mathrm{1}\mid}\right\}.\frac{\mathrm{1}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }} \\ $$$$=\frac{−\mathrm{3}{sign}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}\:\:\:{if}\:{x}\epsilon\left[−\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\right] \\ $$$$=\frac{−\mathrm{3}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }},\frac{\mathrm{3}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}\:\:{ifx}\in\left[−\mathrm{1},\frac{−\mathrm{1}}{\mathrm{2}}\right] \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\mathrm{2}+{x}\right){sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\right)=\left[\left(\mathrm{2}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right){sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\right)\right]_{−\mathrm{1}} ^{\mathrm{1}} −\int_{−\mathrm{1}} ^{−\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{4}{x}+{x}^{\mathrm{2}} }{\mathrm{2}}.\frac{\mathrm{3}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}−\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\mathrm{4}{x}+{x}^{\mathrm{2}} }{\mathrm{2}}.\frac{−\mathrm{3}{dx}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }} \\ $$$${all}\:{we}\:{need}\:{is}\:\int\frac{\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right){dx}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }},{x}={sin}\left(\theta\right),\theta\in\left[−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right]\Rightarrow{cos}\left(\theta\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\int\frac{\mathrm{3}\left({sin}^{\mathrm{2}} +\mathrm{4}{sin}\left(\theta\right)\right){cos}\left(\theta\right){d}\theta}{\left\{\mathrm{2}+{sin}\left(\theta\right)\right\}.{cos}\left(\theta\right).\sqrt{\mathrm{3}}}=\int\frac{\mathrm{3}\left({sin}\left(\theta\right)+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{12}}{\sqrt{\mathrm{3}}\left(\mathrm{2}+{sin}\left(\theta\right)\right\}}{d}\theta \\ $$$$=\sqrt{\mathrm{3}}\int\left({sin}\left(\theta\right)+\mathrm{2}\right)−\mathrm{4}\sqrt{\mathrm{3}}\int\frac{{d}\theta}{\mathrm{2}+{sin}\left(\theta\right)}\:\:\:,{tg}\left(\frac{\theta}{\mathrm{2}}\right)={t}\:{in}\mathrm{2}{nd} \\ $$$$=−\sqrt{\mathrm{3}}{cos}\left(\theta\right)+\mathrm{2}\sqrt{\mathrm{3}}\theta−\mathrm{4}\sqrt{\mathrm{3}}\int\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=−\sqrt{\mathrm{3}}{cos}\left(\theta\right)+\mathrm{2}\sqrt{\mathrm{3}\:\theta}\:−\mathrm{4}\sqrt{\mathrm{3}}\int\frac{{dt}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}} \\ $$$$=−\sqrt{\mathrm{3}}{cos}\left(\theta\right)+\mathrm{2}\sqrt{\mathrm{3}}\theta−\mathrm{8}{arctan}\left(\left(\frac{\mathrm{2}{tg}\left(\frac{\theta}{\mathrm{2}}\right)}{\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right) \\ $$$$\int\frac{\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right){dx}}{\left(\mathrm{2}+{x}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}=−\sqrt{\mathrm{3}}\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\:+\mathrm{2}\sqrt{\mathrm{3}}{arcsin}\left({x}\right)−\mathrm{8}{arctan}\left(\frac{\mathrm{2}{tg}\left(\frac{{arcsin}\left({x}\right)}{\mathrm{2}}\right)+\mathrm{1}}{\sqrt{\mathrm{3}}}\right). \\ $$$$\left.−\left\{−\sqrt{\mathrm{3}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}\frac{\pi}{\mathrm{6}}−\mathrm{8}.\frac{\pi}{\mathrm{12}}.\right)\right\}+\frac{\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{2}\sqrt{\mathrm{3}}.\frac{\pi}{\mathrm{2}}+\mathrm{8}.\frac{\pi}{\mathrm{6}}\right\}+\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}\sqrt{\mathrm{3}}.\frac{\pi}{\mathrm{2}}−\mathrm{8}.\frac{\pi}{\mathrm{3}}\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\pi}{\sqrt{\mathrm{3}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by ajfour last updated on 15/Nov/19
$${Thanks},\:\mathcal{P}{owerful}\:\mathcal{M}{ind}! \\ $$
Commented by mind is power last updated on 15/Nov/19
$${y}\:'{re}\:{welcom} \\ $$
Answered by MJS last updated on 15/Nov/19
![∫(2+x)arcsin ((√(3−3x^2 ))/(2+x)) dx= [t=arccos ((2x+1)/(x+2)) → dx=−(((x+2)(√(3−3x^2 )))/3)dt] (1) for 0≤t≤(π/2) =−9∫((t sin t)/((2−cos t)^3 ))dt= [((∫u′v=uv−∫uv′)),((u′=((sin t)/((2−cos t)^3 )) ⇒ u=−(1/(2(2−cos t)^2 )))),((v=t ⇒ v′=1)) ] =((9t)/(2(2−cos t)^2 ))−(9/2)∫(dt/((2−cos t)^2 ))= ∫(dt/((2−cos t)^2 ))= [w=tan (t/2) → dt=((2dw)/(w^2 +1))] =2∫((w^2 +1)/((3w^2 +1)^2 ))dw= =((2w)/(3(3w^2 +1)))+((4(√3))/9)arctan (√3)w = =((sin t)/(3(2−cos t)))+((4(√3))/9)arctan ((√3)tan (t/2)) =((9t)/(2(2−cos t)^2 ))−((3sin t)/(2(2−cos t)))−2(√3)arctan ((√3)tan (t/2)) (2) for (π/2)≤t≤π =−9π∫((sin t)/((2−cos t)^3 ))dt+9∫((t sin t)/((2−cos t)^3 ))dt= =((9π)/(2(2−cos t)^2 ))−(the above) ⇒ ∫_(−1) ^1 (2+x)arcsin ((√(3−3x^2 ))/(2+x)) dx= =((√3)/3)π+(3/2)](Q73714.png)
$$\int\left(\mathrm{2}+{x}\right)\mathrm{arcsin}\:\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arccos}\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+\mathrm{2}}\:\rightarrow\:{dx}=−\frac{\left({x}+\mathrm{2}\right)\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{3}}{dt}\right] \\ $$$$\left(\mathrm{1}\right) \\ $$$$\mathrm{for}\:\mathrm{0}\leqslant{t}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$=−\mathrm{9}\int\frac{{t}\:\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\begin{bmatrix}{\int{u}'{v}={uv}−\int{uv}'}\\{{u}'=\frac{\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }\:\Rightarrow\:{u}=−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }}\\{{v}={t}\:\Rightarrow\:{v}'=\mathrm{1}}\end{bmatrix} \\ $$$$=\frac{\mathrm{9}{t}}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }−\frac{\mathrm{9}}{\mathrm{2}}\int\frac{{dt}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }= \\ $$$$ \\ $$$$\:\:\:\:\:\int\frac{{dt}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{w}=\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:\rightarrow\:{dt}=\frac{\mathrm{2}{dw}}{{w}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$\:\:\:\:\:=\mathrm{2}\int\frac{{w}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{3}{w}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dw}= \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}{w}}{\mathrm{3}\left(\mathrm{3}{w}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\sqrt{\mathrm{3}}{w}\:= \\ $$$$\:\:\:\:\:=\frac{\mathrm{sin}\:{t}}{\mathrm{3}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)}+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\left(\sqrt{\mathrm{3}}\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$=\frac{\mathrm{9}{t}}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }−\frac{\mathrm{3sin}\:{t}}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)}−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{arctan}\:\left(\sqrt{\mathrm{3}}\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{for}\:\frac{\pi}{\mathrm{2}}\leqslant{t}\leqslant\pi \\ $$$$=−\mathrm{9}\pi\int\frac{\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }{dt}+\mathrm{9}\int\frac{{t}\:\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }{dt}= \\ $$$$=\frac{\mathrm{9}\pi}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }−\left(\mathrm{the}\:\mathrm{above}\right) \\ $$$$ \\ $$$$\Rightarrow \\ $$$$ \\ $$$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{2}+{x}\right)\mathrm{arcsin}\:\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\:{dx}= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\pi+\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 15/Nov/19
$$\mathcal{S}{uperb}\:{Sir};\:{thanks}\:\left({you}\:{are}\right. \\ $$$$\left.{indeed}\:{an}\:{integral}\:{lover}\right)! \\ $$
Commented by ajfour last updated on 15/Nov/19
$${Sir}\:{what}\:{if} \\ $$$${I}=\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{2}+{x}\right)^{\mathrm{2}} \mathrm{arcsin}\:\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\:{dx}\:\:? \\ $$
Commented by MJS last updated on 15/Nov/19
$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{we}\:\mathrm{get} \\ $$$$−\mathrm{27}\int\frac{{t}\:\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }{dt}\:\mathrm{and}\:\left(−\mathrm{27}\int\frac{\left(\pi−{t}\right)\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }{dt}\right) \\ $$$$\mathrm{same}\:\mathrm{path}\:\mathrm{leads}\:\mathrm{to} \\ $$$$−\mathrm{27}\int\frac{{t}\:\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }{dt}= \\ $$$$=\frac{\mathrm{9}{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }+\frac{\mathrm{3sin}\:{t}\:\left(\mathrm{5}−\mathrm{2cos}\:{t}\right)}{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{2}} }+\mathrm{3}\sqrt{\mathrm{3}}\mathrm{arctan}\:\left(\sqrt{\mathrm{3}}\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\right) \\ $$$$−\mathrm{27}\int\frac{\left(\pi−{t}\right)\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }{dt}= \\ $$$$=−\mathrm{27}\pi\int\frac{\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }{dt}+\mathrm{27}\int\frac{{t}\:\mathrm{sin}\:{t}}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{4}} }= \\ $$$$=\frac{\mathrm{9}\pi}{\left(\mathrm{2}−\mathrm{cos}\:{t}\right)^{\mathrm{3}} }−\left(\mathrm{the}\:\mathrm{above}\right) \\ $$$$\Rightarrow \\ $$$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{2}+{x}\right)^{\mathrm{2}} \mathrm{arcsin}\:\frac{\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}+{x}}\:{dx}= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\pi+\frac{\mathrm{15}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 15/Nov/19
$$\mathcal{G}{reat}\:,\:{Sir}.\:{Absolutely}\:{correct}, \\ $$$${I}\:{applied}\:{it},\:{thanks}. \\ $$
Commented by MJS last updated on 15/Nov/19
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$