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dx-x-3-x-2-a-2-




Question Number 118113 by bemath last updated on 15/Oct/20
∫ (dx/(x^3  (√(x^2 −a^2 )))) =?
dxx3x2a2=?
Answered by Lordose last updated on 15/Oct/20
  x=asecu dx= secutanudu  (1/a)∫ ((secutanu)/(sec^3 utanu))du  (1/a)∫(1/(sec^2 u))du  (1/a)∫cos^2 udu  (1/a)((u/2) + ((sinucosu)/2)) + C  secu= (x/a) ⇒ cosu=(a/x)  sinu=(√(1−cos^2 u)) = (√(1−((a/x))^2 ))  (1/(2a))(sec^(−1) ((x/a)) + ((a(√(x^2 −a^2 )))/x^2 )) + C
x=asecudx=secutanudu1asecutanusec3utanudu1a1sec2udu1acos2udu1a(u2+sinucosu2)+Csecu=xacosu=axsinu=1cos2u=1(ax)212a(sec1(xa)+ax2a2x2)+C
Commented by bemath last updated on 15/Oct/20
yes...thank you
yesthankyou
Answered by 1549442205PVT last updated on 15/Oct/20
Put (√(x^2 −a^2 )) =t⇒dt=((xdx)/( (√(x^2 −a^2 ))))  dx=tdt/x  ∫ (dx/(x^3  (√(x^2 −a^2 )))) =∫ ((tdt)/(x^4 .t))=∫(dt/((t^2 +a^2 )^2 ))  I_2 =∫(dt/((t^2 +a^2 )^2 )).Put t=atanϕ  ⇒dt=a(1+tan^2 ϕ)dϕ=a(1+t^2 )dϕ  I_2 =∫((a(1+tan^2 ϕ)dϕ)/(a^4 (tan^2 ϕ+1)^2 ))=∫(dϕ/(a^3 (1+tan^2 ϕ)))  =(1/a^3 )∫cos^2 ϕdϕ=(1/(2a^3 ))∫(1+cos2ϕ)dϕ  =(ϕ/(2a^3 ))+(1/(4a^3 ))sin2ϕ=((tan^(−1) (((√(x^2 −a^2 ))/a)))/(2a^3 ))  +(1/(4a^3 ))×((2a(√(x^2 −a^2 )))/x^2 )  =(1/(2a^3 ))[tan^(−1) (((√(x^2 −a^2 ))/a))+((a(√(x^2 −a^2 )))/x^2 )]+C
Putx2a2=tdt=xdxx2a2dx=tdt/xdxx3x2a2=tdtx4.t=dt(t2+a2)2I2=dt(t2+a2)2.Putt=atanφdt=a(1+tan2φ)dφ=a(1+t2)dφI2=a(1+tan2φ)dφa4(tan2φ+1)2=dφa3(1+tan2φ)=1a3cos2φdφ=12a3(1+cos2φ)dφ=φ2a3+14a3sin2φ=tan1(x2a2a)2a3+14a3×2ax2a2x2=12a3[tan1(x2a2a)+ax2a2x2]+C
Answered by TANMAY PANACEA last updated on 15/Oct/20
t^2 =x^2 −a^2   ∫((xdx)/(x^4 (√(x^2 −a^2 ))))  ∫((tdt)/((t^2 +a^2 )^2 t))→t=atanα  ∫((sec^2 αdα)/(a^4 sec^4 α))=(1/a^2 )∫((1+cos2α)/2)dα  (1/a^2 )×(1/2)(α+((sin2α)/2))+C  (1/(2a^2 ))(tan^− ((t/a))+(1/2)×((2tanα)/(1+tan^2 α)))  =(1/(2a^2 ))tan^(−1) (((√(x^2 −a^2 ))/a))+(1/(2a^2 ))(((t/a)/(1+(t^2 /a^2 ))))+C  =(1/(2a^2 ))×tan^(−1) (((√(x^2 −a^2 ))/a))+(1/(2a^2 ))(((√(x^2 −a^2 ))/a)×(a^2 /x^2 ))+c  =(1/(2a^2 ))tan^(−1) (((√(x^2 −a^2 ))/a) )+(1/(2a^2 ))(((a(√(x^2 −a^2 )))/x^2 ))+c
t2=x2a2xdxx4x2a2tdt(t2+a2)2tt=atanαsec2αdαa4sec4α=1a21+cos2α2dα1a2×12(α+sin2α2)+C12a2(tan(ta)+12×2tanα1+tan2α)=12a2tan1(x2a2a)+12a2(ta1+t2a2)+C=12a2×tan1(x2a2a)+12a2(x2a2a×a2x2)+c=12a2tan1(x2a2a)+12a2(ax2a2x2)+c

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