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Question Number 73689 by ajfour last updated on 14/Nov/19
∫_(−1) ^( 1) (2+x)sin^(−1) (((√(3−3x^2 ))/(2+x)))dx = ?
∫−11(2+x)sin−1(3−3x22+x)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)))
ddx{sin−1(3−3x22+x)}=f(x)′={−3x(2+x)3−3x2−3−3x2(2+x)2}.11−3−3x2(2+x)2={−3x(2+x)−(3−3x2)(2+x)2.3−3x2}.2+x(2x+1)2={−6x−3∣2x+1∣}.1(2+x)3−3x2=−3sign(2x+1)(2+x)3−3x2ifxϵ[−12,1]=−3(2+x)3−3x2,3(2+x)3−3x2ifx∈[−1,−12]∫−11(2+x)sin−1(3−3x22+x)=[(2x+x22)sin−1(3−3x22+x)]−11−∫−1−124x+x22.3(2+x)3−3x2−∫−1214x+x22.−3dx(2+x)3−3x2allweneedis∫3(x2+4x)dx(2+x)3−3x2,x=sin(θ),θ∈[−π2,π2]⇒cos(θ)⩾0⇒∫3(sin2+4sin(θ))cos(θ)dθ{2+sin(θ)}.cos(θ).3=∫3(sin(θ)+2)2−123(2+sin(θ)}dθ=3∫(sin(θ)+2)−43∫dθ2+sin(θ),tg(θ2)=tin2nd=−3cos(θ)+23θ−43∫22+2t1+t2.dt1+t2=−3cos(θ)+23θ−43∫dtt2+t+1=−3cos(θ)+23θ−8arctan((2tg(θ2)3+13))∫(x2+4x)dx(2+x)3−3x2=−31−x2+23arcsin(x)−8arctan(2tg(arcsin(x)2)+13).−{−3.32−23π6−8.π12.)}+12{−23.π2+8.π6}+12{23.π2−8.π3}=32+π3
Commented by ajfour last updated on 15/Nov/19
Thanks, Powerful Mind!
Thanks,PowerfulMind!
Commented by mind is power last updated on 15/Nov/19
y ′re welcom
y′rewelcom
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)
∫(2+x)arcsin3−3x22+xdx=[t=arccos2x+1x+2→dx=−(x+2)3−3x23dt](1)for0⩽t⩽π2=−9∫tsint(2−cost)3dt=[∫u′v=uv−∫uv′u′=sint(2−cost)3⇒u=−12(2−cost)2v=t⇒v′=1]=9t2(2−cost)2−92∫dt(2−cost)2=∫dt(2−cost)2=[w=tant2→dt=2dww2+1]=2∫w2+1(3w2+1)2dw==2w3(3w2+1)+439arctan3w==sint3(2−cost)+439arctan(3tant2)=9t2(2−cost)2−3sint2(2−cost)−23arctan(3tant2)(2)forπ2⩽t⩽π=−9π∫sint(2−cost)3dt+9∫tsint(2−cost)3dt==9π2(2−cost)2−(theabove)⇒∫1−1(2+x)arcsin3−3x22+xdx==33π+32
Commented by ajfour last updated on 15/Nov/19
Superb Sir; thanks (you are indeed an integral lover)!
SuperbSir;thanks(youareindeedanintegrallover)!
Commented by ajfour last updated on 15/Nov/19
Sir what if I=∫_(−1) ^1 (2+x)^2 arcsin ((√(3−3x^2 ))/(2+x)) dx ?
SirwhatifI=∫1−1(2+x)2arcsin3−3x22+xdx?
Commented by MJS last updated on 15/Nov/19
in this case we get −27∫((t sin t)/((2−cos t)^4 ))dt and (−27∫(((π−t)sin t)/((2−cos t)^4 ))dt) same path leads to −27∫((t sin t)/((2−cos t)^4 ))dt= =((9t)/((2−cos t)^3 ))+((3sin t (5−2cos t))/(2(2−cos t)^2 ))+3(√3)arctan ((√3)tan (t/2)) −27∫(((π−t)sin t)/((2−cos t)^4 ))dt= =−27π∫((sin t)/((2−cos t)^4 ))dt+27∫((t sin t)/((2−cos t)^4 ))= =((9π)/((2−cos t)^3 ))−(the above) ⇒ ∫_(−1) ^1 (2+x)^2 arcsin ((√(3−3x^2 ))/(2+x)) dx= =((√3)/2)π+((15)/4)
inthiscaseweget−27∫tsint(2−cost)4dtand(−27∫(π−t)sint(2−cost)4dt)samepathleadsto−27∫tsint(2−cost)4dt==9t(2−cost)3+3sint(5−2cost)2(2−cost)2+33arctan(3tant2)−27∫(π−t)sint(2−cost)4dt==−27π∫sint(2−cost)4dt+27∫tsint(2−cost)4==9π(2−cost)3−(theabove)⇒∫1−1(2+x)2arcsin3−3x22+xdx==32π+154
Commented by ajfour last updated on 15/Nov/19
Great , Sir. Absolutely correct, I applied it, thanks.
Great,Sir.Absolutelycorrect,Iappliedit,thanks.
Commented by MJS last updated on 15/Nov/19
you′re welcome
you′rewelcome

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