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

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Answer 1 · 2018-05-20T13:09:20

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

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

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

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

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,

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

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

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