Using trigonometric substitution what is the integral of #intdx/(x^2sqrt(x^2-1)# ?

Question

#intdx/(x^2sqrt(x^2-1)#

1847 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-08T09:53:34

#I=sqrt(1-1/x^2)+C#

We want to solve

#I=int1/(x^2*sqrt(x^2-1))dx#

Let #x=sec(u)=>dx/(du)=sec(u)tan(u)#

#I=int1/(sec^2(u)*sqrt(sec^2(u)-1))sec(u)tan(u)du#

Use the identity #sec^2(a)-1=tan^2(a)#

#I=int1/(sec^2(u)*tan(u))sec(u)tan(u)du#

#=int1/sec(u)du#

#=intcos(u)du#

#=sin(u)+C#

Substitute #u=sec^-1(x)#

#I=sin(sec^-1(x))+C#

Use #sin(sec^-1(x))=sqrt(1-1/x^2)#

#I=sqrt(1-1/x^2)+C#

Answer 2 · 2018-02-08T12:59:47

#int (dx)/[x^2*sqrt(x^2-1)]=sqrt(x^2-1)/x+C#

#int (dx)/[x^2*sqrt(x^2-1)]#

After using #x=coshu# and #dx=sinhu*du# transforms, this integral became

#int (sinhu*du)/[(coshu)^2*sqrt((coshu)^2-1)]#

=#int (sinhu*du)/[(coshu)^2*sqrt((sinhu)^2)]#

=#int (sinhu*du)/[(coshu)^2*sinhu]#

=#int (sechu)^2*du#

=#tanhu+C#

=#sinhu/coshu+C#

=#sqrt((coshu)^2-1)/coshu+C#

=#sqrt(x^2-1)/x+C#