How do you integrate $intxsec^-1(x)dx$ ?

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Answer 1 · 2018-04-10T07:55:44

The answer is $=x^2/2arcsecx-1/2sqrt(x^2-1)+C$

$intuv'=uv-intu'v$

Here,

$u=arcsecx$ , $=>$ , $u'=(dx)/(xsqrt(x^2-1))$

$v'=x$ , $=>$ , $v=x^2/2$

Therefore,

$intxarcsecxdx=x^2/2arcsecx-1/2int(xdx)/(sqrt(x^2-1))$

Let $u=x^2-1$ , $=>$ , $du=2xdx$

Therefore,

$1/2int(xdx)/(sqrt(x^2-1))=1/4int(du)/(sqrtu)$

$=1/2sqrtu$

$=1/2sqrt(x^2-1)$

And finally,

$intxarcsecxdx=x^2/2arcsecx-1/2sqrt(x^2-1)+C$

How do you integrate intxsec^-1(x)dxintxsec^-1(x)dx?