Question

How do you evaluate the integral `int x"arcsec"(x^2)`?

Answer 1

`int x*arcsec(x^2)*dx`

=`1/2x^2*arcsec(x^2)-1/2ln(x^2+sqrt(x^4-1))+C`

Explanation:

`int x*arcsec(x^2)*dx`

=`1/2int 2x*arcsec(x^2)*dx`

After using `y=x^2` and `2x*dx=dy` transforms, this integral became

`1/2int arcsecy*dy`

After using `z=arcsecy`, `y=secz` and `dy=secz*tanz*dz` transforms, it became

`1/2int z*secz*tanz*dz`

=`1/2z*secz-1/2int secz*dz`

=`1/2z*secz-1/2int (secz*(secz+tanz)*dz)/(secz+tanz)`

=`1/2z*secz-1/2ln(secz+tanz)+C`

For `y=secz`, `tanz` must be equal to `sqrt(y^2-1)`. Thus,

`1/2y*arcsecy-1/2ln(y+sqrt(y^2-1))+C`

=`1/2x^2*arcsec(x^2)-1/2ln(x^2+sqrt(x^4-1))+C`