Question

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

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

Answer 1

`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`

Answer 2

`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`