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

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

2 Answers
Feb 8, 2018

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

Explanation:

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

Feb 8, 2018

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

Explanation:

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