Question Number 118113 by bemath last updated on 15/Oct/20

Answered by Lordose last updated on 15/Oct/20

Commented by bemath last updated on 15/Oct/20

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](https://www.tinkutara.com/question/Q118126.png)
Answered by TANMAY PANACEA last updated on 15/Oct/20
