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3sin-x-2-2sin-x-3-2-dx-




Question Number 137249 by bemath last updated on 31/Mar/21
∫ (((3sin x+2))/((2sin x+3)^2 )) dx =?
(3sinx+2)(2sinx+3)2dx=?
Answered by bemath last updated on 31/Mar/21
let tan ((x/2))=u   { ((dx=(2/(u^2 +1)) du)),((sin x=((2u)/(u^2 +1)))) :}  I=∫ (((((6u+2u^2 +2)/(u^2 +1))))/((((4u+3u^2 +3)/(u^2 +1)))^2 )) ((2/(u^2 +1)))du  I= ∫ ((4(u^2 +3u+1))/((3u^2 +4u+3)^2 )) du  ....
lettan(x2)=u{dx=2u2+1dusinx=2uu2+1I=(6u+2u2+2u2+1)(4u+3u2+3u2+1)2(2u2+1)duI=4(u2+3u+1)(3u2+4u+3)2du.
Answered by Ar Brandon last updated on 31/Mar/21
I=∫((3sinx+2)/((2sinx+3)^2 ))dx=(3/2)∫((2sinx+3−(5/3))/((2sinx+3)^2 ))dx     =(3/2)∫(dx/(2sinx+3))−(5/2)∫(dx/((2sinx+3)^2 ))      =3∫(dt/((2((2t)/(1+t^2 ))+3)(t^2 +1)))−5∫(dt/((2((2t)/(1+t^2 ))+3)^2 (t^2 +1)))     =3∫(dt/(3t^2 +4t+3))−5∫((t^2 +1)/((3t^2 +4t+3)^2 ))dt  f(t)=((t^2 +1)/((3t^2 +4t+3)^2 ))=(1/3)(((3t^2 +4t+3−4t)/((3t^2 +4t+3)^2 )))        =(1/(3(3t^2 +4t+3)))−((4t)/(3(3t^2 +4t+3)^2 ))  g(t)=((4t)/(3(3t^2 +4t+3)^2 ))=(1/3){(2/3)∙((6t+4)/((3t^2 +4t+3)^2 ))−(8/3)∙(1/((3t^2 +4t+3)^2 ))}
I=3sinx+2(2sinx+3)2dx=322sinx+353(2sinx+3)2dx=32dx2sinx+352dx(2sinx+3)2=3dt(22t1+t2+3)(t2+1)5dt(22t1+t2+3)2(t2+1)=3dt3t2+4t+35t2+1(3t2+4t+3)2dtf(t)=t2+1(3t2+4t+3)2=13(3t2+4t+34t(3t2+4t+3)2)=13(3t2+4t+3)4t3(3t2+4t+3)2g(t)=4t3(3t2+4t+3)2=13{236t+4(3t2+4t+3)2831(3t2+4t+3)2}
Answered by Dwaipayan Shikari last updated on 31/Mar/21
∫((asinx+b)/((bsinx+a)^2 ))dx=τ(a,b)  =(a^2 /b)∫((absin(x)+a^2 )/((absin(x)+a^2 )^2 ))+((b^2 −a^2 )/((absin(x)+a^2 )^2 ))dx  =(a/b^2 )∫(1/(sin(x)+(a/b)))dx+(((b^2 −a^2 ))/b^3 )∫(1/((sin(x)+(a/b))^2 ))dx     tan(x/2)=t  J(a)=∫(1/(sin(x)+(a/b)))dx=2∫(1/(((2t)/(1+t^2 ))+(a/b))).(1/(1+t^2 ))dt  =((2b)/a)∫(1/(t^2 +2(b/a)t+1))dt=((2b)/a)∫(1/((t+(√(b/a)))^2 +((b−a)/a)))dt  =((2b)/a)(√(a/(b−a)))tan^(−1) (((√a)t+(√b))/( (√(b−a))))+C  J′(a)=−(1/b)∫(1/((sinx+(a/b))^2 ))dx  J′(a)=tan^(−1) (((√a)t+(√b))/( (√(b−a))))(−((b(b−a)^(−1/2) )/a^(3/2) )+(a^(−1/2) /((b−a)^(3/2) )))+((b−a)/(b−a+((√a)t+(√b))^2 ))((((√a)t+b)/( 2(b−a)^(3/2) ))+((t/( (√a)))/2))    ...τ(a,b)=(a^2 /b)J(a)−((b^2 −a^2 )/b^4 )J′(a)
asinx+b(bsinx+a)2dx=τ(a,b)=a2babsin(x)+a2(absin(x)+a2)2+b2a2(absin(x)+a2)2dx=ab21sin(x)+abdx+(b2a2)b31(sin(x)+ab)2dxtanx2=tJ(a)=1sin(x)+abdx=212t1+t2+ab.11+t2dt=2ba1t2+2bat+1dt=2ba1(t+ba)2+baadt=2baabatan1at+bba+CJ(a)=1b1(sinx+ab)2dxJ(a)=tan1at+bba(b(ba)1/2a3/2+a1/2(ba)3/2)+baba+(at+b)2(at+b2(ba)3/2+ta2)τ(a,b)=a2bJ(a)b2a2b4J(a)
Answered by MJS_new last updated on 31/Mar/21
(u/v)′=((u′v−v′u)/v^2 ) ⇔ (u/v)=∫((u′)/v)−((v′u)/v^2 ) ⇔ ∫((v′u)/v^2 )=−(u/v)+∫((u′)/v)  v=3+2sin x ⇒ v′=2cos x  2ucos x =2+3sin x ⇒ u=((2+3sin x)/(2cos x)) ⇒ u′=((3+2sin x)/(2cos^2  x))  now we have  ∫((v′u)/v^2 )=−(u/v)+∫((u′)/v)  ∫((2+3sin x)/((3+2sin x)^2 ))dx=−((2+3sin x)/(2(3+2sin x)cos x))+(1/2)∫(dx/(cos^2  x))=  =−((2+3sin x)/(2(3+2sin x)cos x))+(1/2)tan x =  =−((cos x)/(3+2sin x))+C
(u/v)=uvvuv2uv=uvvuv2vuv2=uv+uvv=3+2sinxv=2cosx2ucosx=2+3sinxu=2+3sinx2cosxu=3+2sinx2cos2xnowwehavevuv2=uv+uv2+3sinx(3+2sinx)2dx=2+3sinx2(3+2sinx)cosx+12dxcos2x==2+3sinx2(3+2sinx)cosx+12tanx==cosx3+2sinx+C
Commented by Dwaipayan Shikari last updated on 31/Mar/21
So it has been proved that experience pays more :(
Soithasbeenprovedthatexperiencepaysmore:(
Commented by MJS_new last updated on 31/Mar/21
how old are you?  I′ve been solving integrals for about 35 years...  this special one I tried using t=tan (x/2) first;  then I found Ostrogradski′s Method leads to  ((p_1 (t))/(q_1 (t)))+∫0dt so I knew it can be solved by parts  using the quotient rule most people don′t  know...
howoldareyou?Ivebeensolvingintegralsforabout35yearsthisspecialoneItriedusingt=tanx2first;thenIfoundOstrogradskisMethodleadstop1(t)q1(t)+0dtsoIknewitcanbesolvedbypartsusingthequotientrulemostpeopledontknow
Commented by Dwaipayan Shikari last updated on 31/Mar/21
e^((17)/6)   (Exactly maybe)  :)
e176(Exactlymaybe):)
Commented by Ar Brandon last updated on 31/Mar/21
35 years ! That′s the sum of our ages   😅
35years!Thatsthesumofourages😅
Answered by EDWIN88 last updated on 31/Mar/21
E = ∫ (((2+3sin x))/((3+2sin x)^2 )) dx   E = ∫ ((2+3sin x)/(2cos x)). ((2cos x)/((3+2sin x)^2 )) dx  E = (1/2)∫ ((2+3sin x)/(cos x))_(u)  .((2cos x)/((3+2sin x)^2 )).dx_(dv)   by parts   ⇒du = ((3+2sin x)/(cos^2 x)) dx ; v = −(1/(3+2sin x))  2E=−((2+3sin x)/(cos x(3+2sin x)))+∫ (dx/(cos^2 x))   2E = −((2+3sin x)/(cos x(3+2sin x))) + tan x + c   E = −((2+3sin x)/(2cos x(3+2sin x)))+ ((sin x)/(2cos x))+ C  E=((−2−3sin x+sin x(3+2sin x))/(2cos x(3+2sin x))) + C  E = ((2sin^2 x−2)/(2cos x(3+2sin x))) = ((−cos x)/(3+2sin x)) + C
E=(2+3sinx)(3+2sinx)2dxE=2+3sinx2cosx.2cosx(3+2sinx)2dxE=122+3sinxcosxu.2cosx(3+2sinx)2.dxdvbypartsdu=3+2sinxcos2xdx;v=13+2sinx2E=2+3sinxcosx(3+2sinx)+dxcos2x2E=2+3sinxcosx(3+2sinx)+tanx+cE=2+3sinx2cosx(3+2sinx)+sinx2cosx+CE=23sinx+sinx(3+2sinx)2cosx(3+2sinx)+CE=2sin2x22cosx(3+2sinx)=cosx3+2sinx+C

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